算法笔记--CDQ分治 && 整体二分

参考:https://www.luogu.org/blog/Owencodeisking/post-xue-xi-bi-ji-cdq-fen-zhi-hu-zheng-ti-er-fen

前置技能：树状数组，线段树，分治、归并排序

CDQ分治：

据说是OI大佬陈丹琦发明的

1.三维偏序

思路：

第一维排序，第二维分治，第三维树状数组上查询

考虑分治时区间 [l, m] 对区间 [m+1, r] 的贡献，因为第一维已经排好序，所以区间 [l, m] 的第一维小于区间 [m+1, r]的第一维

然后对于区间 [m+1, r]中的某个元素x，将区间 [l, m] 的第二维小于x的元素的按第三维的权值加入树状数组，

最后区间 [l, m] 对区间 x 的贡献就是查询树状数组中小于x第三维的个数

可以边进行分治边进行归并排序，树状数组要及时清空

通过画图我们可以发现，对于每个位置，我们在分治时，它之前的位置对它的贡献都计算过了，所以这种方法是正确的。

因为递归的层数是log(n)层，再加上树状数组，所以时间复杂度是O(n*log(n)^2)

P3810 【模板】三维偏序（陌上花开）

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + , M = 2e5 + ;

struct Node {

    int x, y, z;

    int ans, cnt;

    bool operator < (const Node & rhs) const {

        if(x == rhs.x) {

            if(y == rhs.y) return z < rhs.z;

            else return y < rhs.y;

        }

        else return x < rhs.x;

    }

}a[N], tmp[N];

int bit[M], res[N], n, k, cnt = ;

void add(int x, int a) {

    while(x <= k) bit[x] += a, x += x&-x;

}

int sum(int x) {

    int res = ;

    while(x) res += bit[x], x -= x&-x;

    return res;

}

void cdq(int l, int r) {

    if(l == r) {

        a[l].ans += a[l].cnt-;

        return ;

    }

    int m = l+r >> ;

    cdq(l, m);

    cdq(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && a[p].y <= a[q].y) add(a[p].z, a[p].cnt), tmp[tp++] = a[p], ++p;

        a[q].ans += sum(a[q].z);

        tmp[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i) add(a[i].z, -a[i].cnt);

    while(p <= m) tmp[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = tmp[i];

}

int main() {

    scanf("%d %d", &n, &k);

    for (int i = ; i <= n; ++i) scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z);

    sort(a+, a++n);

    int now = ;

    for (int i = ; i <= n; ++i) {

        if(a[i].x == a[i-].x && a[i].y == a[i-].y && a[i].z == a[i-].z) ++now;

        else {

            a[++cnt] = a[i-];

            a[cnt].cnt = now;

            a[cnt].ans = ;

            now = ;

        }

    }

    a[++cnt] = a[n];

    a[cnt].cnt = now;

    a[cnt].ans = ;

    cdq(, cnt);

    for (int i = ; i <= cnt; ++i) res[a[i].ans] += a[i].cnt;

    for (int i = ; i < n; ++i) printf("%d\n", res[i]);

    return ;

}

例题1：CodeForces - 669E

思路：时间看成一个维度就转换成了三维偏序了

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + ;

struct node {

    int a, t, x, ans, id;

    bool operator < (const node & rhs) const {

        return id < rhs.id;

    }

}a[N], tmp[N];

int n;

map<int, int> cnt;

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m);

    cdq(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && a[p].t <= a[q].t) {

            if(a[p].a == ) cnt[a[p].x]++;

            else if(a[p].a == ) cnt[a[p].x]--;

            tmp[tp++] = a[p], ++p;

        }

        if(a[q].a == ) a[q].ans += cnt[a[q].x];

        tmp[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i)  {

        if(a[i].a == ) cnt[a[i].x]--;

        else if(a[i].a == ) cnt[a[i].x]++;

    }

    while(p <= m) tmp[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = tmp[i];

}

int main() {

    scanf("%d", &n);

    for (int i = ; i <= n; ++i) scanf("%d %d %d", &a[i].a, &a[i].t, &a[i].x), a[i].ans = , a[i].id = i;

    cdq(, n);

    sort(a+, a++n);

    for (int i = ; i <= n; ++i) if(a[i].a == ) printf("%d\n", a[i].ans);

    return ;

}

例题2：HDU - 5618

思路：由于对于每个点都要询问，所以不能像陌上花开那样缩点了，排序后把相同的点的贡献先加上去

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + ;

struct Node {

    int x, y, z, id;

    int cnt;

    bool operator < (const Node & rhs) const {

        if(x == rhs.x) {

            if(y == rhs.y) return z < rhs.z;

            else return y < rhs.y;

        }

        else return x < rhs.x;

    }

}a[N], tmp[N];

int bit[N], ans[N], n, cnt = ;

void add(int x, int a) {

    while(x < N) bit[x] += a, x += x&-x;

}

int sum(int x) {

    int res = ;

    while(x) res += bit[x], x -= x&-x;

    return res;

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m);

    cdq(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && a[p].y <= a[q].y) add(a[p].z, a[p].cnt), tmp[tp++] = a[p], ++p;

        ans[a[q].id] += sum(a[q].z);

        tmp[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i) add(a[i].z, -a[i].cnt);

    while(p <= m) tmp[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = tmp[i];

}

int T;

int main() {

    scanf("%d", &T);

    while(T--) {

        scanf("%d", &n);

        for (int i = ; i <= n; ++i) scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z), a[i].id = i, a[i].cnt = , ans[i] = ;

        sort(a+, a++n);

        int now = ;

        for (int i = n-; i >= ; --i) {

            if(a[i].x == a[i+].x && a[i].y == a[i+].y && a[i].z == a[i+].z) ++now;

            else now = ;

            ans[a[i].id] += now;

        }

        cdq(, n);

        for (int i = ; i <= n; ++i) printf("%d\n", ans[i]);

    }

    return ;

}

例题3：CodeChef - QRECT

思路：考虑用容斥，用总个数减去和横纵坐标和它不相交的个数，这样我们发现和它横纵坐标都不相交被减了两次，也就是四个角上的矩形，四个角上的矩形的个数就是三维偏序问题

代码：

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define pb push_back

//head

const int N = 1e5 + , M = 2e5 + ;

struct Node {

    int ty, x1, y1, x2, y2, id;

    int ans, cnt;

    bool operator < (const Node & rhs) const {

        return id < rhs.id;

    }

}a[N], aa[N], tmp[N];

vector<int> vx, vy;

int n, bit1[M], bit2[M], bit[M], p, pos[N], cnt = , now = ;

char op[];

void add(int x, int a, int *bit) {

    while(x < M) bit[x] += a, x += x&-x;

}

int sum(int x, int *bit) {

    int res = ;

    while(x) res += bit[x], x -= x&-x;

    return res;

}

void init() {

    for (int i = ; i < M; ++i) bit1[i] = bit2[i] = ;

}

void cdq1(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq1(l, m);

    cdq1(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && aa[p].x1 < aa[q].x1) {

            if(aa[p].ty == ) add(aa[p].y1, , bit);

            else if(aa[p].ty == ) add(aa[p].y1, -, bit);

            tmp[tp++] = aa[p];

            ++p;

        }

        if(aa[q].ty == ) {

            aa[q].ans += sum(aa[q].y1-, bit);

        }

        tmp[tp++] = aa[q];

        ++q;

    }

    for (int i = l; i < p; ++i) if(aa[i].ty == ) add(aa[i].y1, -, bit); else if(aa[i].ty == ) add(aa[i].y1, , bit);

    while(p <= m) tmp[tp++] = aa[p], ++p;

    for (int i = l; i <= r; ++i) aa[i] = tmp[i];

}

void cdq2(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq2(l, m);

    cdq2(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && aa[p].x1 > aa[q].x1) {

            if(aa[p].ty == ) add(aa[p].y1, , bit);

            else if(aa[p].ty == ) add(aa[p].y1, -, bit);

            tmp[tp++] = aa[p];

            ++p;

        }

        if(aa[q].ty == ) {

            aa[q].ans += sum(M-, bit) - sum(aa[q].y1, bit);

        }

        tmp[tp++] = aa[q];

        ++q;

    }

    for (int i = l; i < p; ++i) if(aa[i].ty == ) add(aa[i].y1, -, bit); else if(aa[i].ty == ) add(aa[i].y1, , bit);

    while(p <= m) tmp[tp++] = aa[p], ++p;

    for (int i = l; i <= r; ++i) aa[i] = tmp[i];

}

void cdq3(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq3(l, m);

    cdq3(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && aa[p].x1 < aa[q].x1) {

            if(aa[p].ty == ) add(aa[p].y1, , bit);

            else if(aa[p].ty == ) add(aa[p].y1, -, bit);

            tmp[tp++] = aa[p];

            ++p;

        }

        if(aa[q].ty == ) {

            aa[q].ans += sum(M-, bit) - sum(aa[q].y1, bit);

        }

        tmp[tp++] = aa[q];

        ++q;

    }

    for (int i = l; i < p; ++i) if(aa[i].ty == ) add(aa[i].y1, -, bit); else if(aa[i].ty == ) add(aa[i].y1, , bit);

    while(p <= m) tmp[tp++] = aa[p], ++p;

    for (int i = l; i <= r; ++i) aa[i] = tmp[i];

}

void cdq4(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq4(l, m);

    cdq4(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && aa[p].x1 > aa[q].x1) {

            if(aa[p].ty == ) add(aa[p].y1, , bit);

            else if(aa[p].ty == ) add(aa[p].y1, -, bit);

            tmp[tp++] = aa[p];

            ++p;

        }

        if(aa[q].ty == ) {

            aa[q].ans += sum(aa[q].y1-, bit);

        }

        tmp[tp++] = aa[q];

        ++q;

    }

    for (int i = l; i < p; ++i) if(aa[i].ty == ) add(aa[i].y1, -, bit); else if(aa[i].ty == ) add(aa[i].y1, , bit);

    while(p <= m) tmp[tp++] = aa[p], ++p;

    for (int i = l; i <= r; ++i) aa[i] = tmp[i];

}

int main() {

    scanf("%d", &n);

    for (int i = ; i <= n; ++i) {

        scanf("%s", op);

        if(op[] == 'I') {

            scanf("%d %d %d %d", &a[i].x1, &a[i].y1, &a[i].x2, &a[i].y2);

            if(a[i].x1 > a[i].x2) swap(a[i].x1, a[i].x2);

            if(a[i].y1 > a[i].y2) swap(a[i].y1, a[i].y2);

            vx.pb(a[i].x1);vx.pb(a[i].x2);vy.pb(a[i].y1);vy.pb(a[i].y2);

            a[i].ty = ;

            a[i].id = i;

            pos[++now] = i;

            ++cnt;

        }

        else if(op[] == 'D') {

            scanf("%d", &p);

            a[i] = a[pos[p]];

            a[i].ty = ;

            a[i].id = i;

            --cnt;

        }

        else {

            scanf("%d %d %d %d", &a[i].x1, &a[i].y1, &a[i].x2, &a[i].y2);

            if(a[i].x1 > a[i].x2) swap(a[i].x1, a[i].x2);

            if(a[i].y1 > a[i].y2) swap(a[i].y1, a[i].y2);

            vx.pb(a[i].x1);vx.pb(a[i].x2);vy.pb(a[i].y1);vy.pb(a[i].y2);

            a[i].ty = ;

            a[i].id = i;

            a[i].cnt = a[i].ans = cnt;

        }

    }

    sort(vx.begin(), vx.end());

    vx.erase(unique(vx.begin(), vx.end()), vx.end());

    sort(vy.begin(), vy.end());

    vy.erase(unique(vy.begin(), vy.end()), vy.end());

    for (int i = ; i <= n; ++i) {

        a[i].x1 = lower_bound(vx.begin(), vx.end(), a[i].x1) - vx.begin() + ;

        a[i].x2 = lower_bound(vx.begin(), vx.end(), a[i].x2) - vx.begin() + ;

        a[i].y1 = lower_bound(vy.begin(), vy.end(), a[i].y1) - vy.begin() + ;

        a[i].y2 = lower_bound(vy.begin(), vy.end(), a[i].y2) - vy.begin() + ;

    }

    for (int i = ; i <= n; ++i) {

        if(a[i].ty == ) add(a[i].x1, , bit1), add(a[i].x2, , bit2);

        else if(a[i].ty == ) add(a[i].x1, -, bit1), add(a[i].x2, -, bit2);

        else a[i].ans -= sum(a[i].x1-, bit2) + (sum(M-, bit1) - sum(a[i].x2, bit1));

    }

    init();

    for (int i = ; i <= n; ++i) {

        if(a[i].ty == ) add(a[i].y1, , bit1), add(a[i].y2, , bit2);

        else if(a[i].ty == ) add(a[i].y1, -, bit1), add(a[i].y2, -, bit2);

        else a[i].ans -= sum(a[i].y1-, bit2) + (sum(M-, bit1) - sum(a[i].y2, bit1));

    }

    for (int i = ; i <= n; ++i) {

        aa[i] = a[i];

        if(a[i].ty <= ) aa[i].x1 = a[i].x2, aa[i].y1 = a[i].y2;

    }

    cdq1(, n);

    sort(aa+, aa++n);

    for (int i = ; i <= n; ++i) a[i].ans = aa[i].ans;

    for (int i = ; i <= n; ++i) {

        aa[i] = a[i];

        if(a[i].ty == ) aa[i].x1 = a[i].x2, aa[i].y1 = a[i].y2;

    }

    cdq2(, n);

    sort(aa+, aa++n);

    for (int i = ; i <= n; ++i) a[i].ans = aa[i].ans;

    for (int i = ; i <= n; ++i) {

        aa[i] = a[i];

        if(a[i].ty <= ) aa[i].x1 = a[i].x2, aa[i].y1 = a[i].y1;

        else aa[i].x1 = a[i].x1, aa[i].y1 = a[i].y2;

    }

    cdq3(, n);

    sort(aa+, aa++n);

    for (int i = ; i <= n; ++i) a[i].ans = aa[i].ans;

    for (int i = ; i <= n; ++i) {

        aa[i] = a[i];

        if(a[i].ty <= ) aa[i].x1 = a[i].x1, aa[i].y1 = a[i].y2;

        else aa[i].x1 = a[i].x2, aa[i].y1 = a[i].y1;

    }

    cdq4(, n);

    sort(aa+, aa++n);

    for (int i = ; i <= n; ++i) a[i].ans = aa[i].ans;

    for (int i = ; i <= n; ++i) if(a[i].ty == ) printf("%d\n", a[i].ans);

    return ;

}

例题4：P3157 [CQOI2011]动态逆序对

思路：将删除的时间看成一个维度

代码：

#include<bits/stdc++.h>

using namespace std;

#define LL long long

const int N = 1e5 + ;

struct Node {

    int t, v, pos, id;

    bool operator < (const Node & rhs) const {

        return t < rhs.t;

    }

}a[N], tmp[N];

int pos[N], ans[N], n, m, b;

LL res = ;

struct BIT {

    int bit[N];

    void init() {

        for (int i = ; i <= n; ++i) bit[i] = ;

    }

    void add(int x, int a) {

        while(x <= n) bit[x] += a, x += x&-x;

    }

    int sum(int x) {

        int res = ;

        while(x) res += bit[x], x -= x&-x;

        return res;

    }

}B;

void cdq1(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq1(l, m); cdq1(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && a[p].pos < a[q].pos) {

            B.add(a[p].v, );

            tmp[tp++] = a[p];

            ++p;

        }

        ans[a[q].id] += B.sum(n) - B.sum(a[q].v);

        tmp[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i) B.add(a[i].v, -);

    while(p <= m) tmp[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = tmp[i];

}

void cdq2(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq2(l, m); cdq2(m+, r);

    int p = l, q = m+, tp = l;

    while(q <= r) {

        while(p <= m && a[p].pos > a[q].pos) {

            B.add(a[p].v, );

            tmp[tp++] = a[p];

            ++p;

        }

        ans[a[q].id] += B.sum(a[q].v-);

        tmp[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i) B.add(a[i].v, -);

    while(p <= m) tmp[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = tmp[i];

}

int main() {

    scanf("%d %d", &n, &m);

    for (int i = ; i <= n; ++i) {

        scanf("%d", &a[i].v);

        a[i].pos = i;

        a[i].id = ;

        a[i].t = ;

        pos[a[i].v] = i;

    }

    for (int i = ; i <= n; ++i) {

        res += B.sum(n) - B.sum(a[i].v);

        B.add(a[i].v, );

    }

    B.init();

    for (int i = ; i <= m; ++i) {

        scanf("%d", &b);

        a[pos[b]].id = i;

        a[pos[b]].t = m-i+;

    }

    sort(a+, a++n);

    cdq1(, n);

    sort(a+, a++n);

    cdq2(, n);

    for (int i = ; i <= m; ++i) {

        printf("%lld\n", res);

        res -= ans[i];

    }

    return ;

}

例题5：UVALive - 6667

思路：严格三维偏序，在对x排序时，如果x相同，按y从大到小排序，这样x相同是左边区间就不会对右边区间产生贡献了，因为这时左边y大于等于右边y

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 3e5 + , MM = 1e6 + ;

struct Node {

    int x, y, z, len;

}a[N];

struct BIT {

    int bit[MM];

    void clr(int x) {

        while(x < MM) bit[x] = , x += x&-x;

    }

    void add(int x, int a) {

        while(x < MM) bit[x] = max(bit[x], a), x += x&-x;

    }

    int mx(int x) {

        int res = ;

        while(x) res = max(res, bit[x]), x -= x&-x;

        return res;

    }

}b;

bool cmp1(Node a, Node b) {

    if(a.x == b.x) return a.y > b.y;

    else return a.x < b.x;

}

bool cmp2(Node a, Node b) {

    return a.y < b.y;

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m);

    sort(a+l, a+m+, cmp2);

    sort(a+m+, a+r+, cmp2);

    int p = l, q = m+;

    while(q <= r) {

        while(p <= m && a[p].y < a[q].y) b.add(a[p].z, a[p].len), ++p;

        a[q].len = max(a[q].len, b.mx(a[q].z-)+);

        ++q;

    }

    for (int i = l; i < p; ++i) b.clr(a[i].z);

    sort(a+m+, a+r+, cmp1);

    cdq(m+, r);

}

int A, B, C = ~(<<), M = (<<)-, m, n;

int r() {

    A =  * (A & M) + (A >> );

    B =  * (B & M) + (B >> );

    return (C & ((A << ) + B)) % ;

}

int main() {

    while(~scanf("%d %d %d %d", &m, &n, &A, &B)) {

        if(!m && !n && !A && !B) break;

        for (int i = ; i <= m; ++i) {

            scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z);

            a[i].z++;

            a[i].len = ;

        }

        for (int i = ; i <= n; ++i) {

            ++m;

            a[m].x = r();

            a[m].y = r();

            a[m].z = r();

            a[m].z++;

            a[m].len = ;

        }

        sort(a+, a++m, cmp1);

        cdq(, m);

        int ans = ;

        for (int i = ; i <= m; ++i) ans = max(ans, a[i].len);

        printf("%d\n", ans);

    }

    return ;

}

2.四维偏序

参考：http://www.cnblogs.com/candy99/p/6442434.html

cdq套cdq

假设四维为（a, b, c, d），在进行普通的cdq分治时，我们归并排序使得b有序，这个时候就可以再套一个cdq来解决(b, c, d)的三维偏序问题，但是这时不要忘记了a的作用，

我们要求a有序时左边[l, m]对右边[m+1, r]的贡献，所以在套之前给左边[l, m]的元素打个标记来区分，然后就可以愉快地套cdq了。时间复杂度显然为O(n*log(n)^3)。

然后原题找不到了，做一道稍微复杂点的，不要忘记第二个cdq是在新数组上进行，如果在原数组上进行的话，做完以后b就不是有序的了。

HDU - 5126

思路：将一个查询用容斥拆分成8个查询

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 5e4 + ;

struct Node {

    int ty, x, y, z, c, id;

    bool flag;

}a[N*], t1[N*], t2[N*];

vector<int> vc;

int T, n, ty, x1, y1, z1, x2, y2, z2, tot = , ans[N], cnt = , sz;

struct BIT {

    int bit[N*];

    void add(int x, int a) {

        while(x <= sz) bit[x] += a, x += x&-x;

    }

    int sum(int x) {

        int res = ;

        while(x) res += bit[x], x -= x&-x;

        return res;

    }

}B;

void cdq1(int l, int r){

    if(l == r) return ;

    int m = l+r >> ;

    cdq1(l, m); cdq1(m+, r);

    int p = l, q = m+, tp = l;

    Node *a = t1, *t = t2;

    while(q <= r) {

        while(p <= m && a[p].y <= a[q].y) {

            if(a[p].flag && a[p].ty == ) B.add(a[p].z, );

            t[tp++] = a[p];

            ++p;

        }

        if(!a[q].flag && a[q].ty == ) ans[a[q].id] += a[q].c*B.sum(a[q].z);

        t[tp++] = a[q];

        ++q;

    }

    for (int i = l; i < p; ++i) if(a[i].flag && a[i].ty == ) B.add(a[i].z, -);

    while(p <= m) t[tp++] = a[p], ++p;

    for (int i = l; i <= r; ++i) a[i] = t[i];

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m); cdq(m+, r);

    int p = l, q = m+, tp = l;

    Node *t = t1;

    while(q <= r) {

        while(p <= m && a[p].x <= a[q].x) t[tp] = a[p], t[tp].flag = , ++tp, ++p;

        t[tp] = a[q];

        t[tp].flag = ;

        ++tp;

        ++q;

    }

    while(p <= m) t[tp] = a[p], t[tp].flag = , ++tp, ++p;

    for (int i = l; i <= r; ++i) a[i] = t[i];

    cdq1(l, r);

}

int main() {

    scanf("%d", &T);

    while(T--) {

        scanf("%d", &n);

        tot = cnt = ;

        for (int i = ; i <= n; ++i) {

            scanf("%d", &ty);

            if(ty == ) {

                ++tot;

                a[tot].ty = ;

                scanf("%d %d %d", &a[tot].x, &a[tot].y, &a[tot].z);

            }

            else {

                scanf("%d %d %d %d %d %d", &x1, &y1, &z1, &x2, &y2, &z2);

                ++cnt;

                ans[cnt] = ;

                a[++tot] = {, x2, y2, z2, , cnt, };

                a[++tot] = {, x1-, y2, z2, -, cnt, };

                a[++tot] = {, x2, y1-, z2, -, cnt, };

                a[++tot] = {, x2, y2, z1-, -, cnt, };

                a[++tot] = {, x1-, y1-, z2, , cnt, };

                a[++tot] = {, x1-, y2, z1-, , cnt, };

                a[++tot] = {, x2, y1-, z1-, , cnt, };

                a[++tot] = {, x1-, y1-, z1-, -, cnt, };

            }

        }

        vc.clear();

        for (int i = ; i <= tot; ++i) vc.pb(a[i].z);

        sort(vc.begin(), vc.end());

        vc.erase(unique(vc.begin(), vc.end()), vc.end());

        for (int i = ; i <= tot; ++i) a[i].z = lower_bound(vc.begin(), vc.end(), a[i].z) - vc.begin()+;

        sz = (int)vc.size();

        cdq(, tot);

        for (int i = ; i <= cnt; ++i) printf("%d\n", ans[i]);

    }

    return ;

}

3.其他

cdq优化dp等。

例题1：HYSBZ - 4553

思路：设mx[i]为第i个的最大值，mn[i]为最小值

dp[i] = max{dp[j] + 1} 其中 j < i, mx[j] <= a[i], a[j] <= mn[i]

这是个三维偏序问题，可以用cdq维护dp的转移，具体看代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + ;

struct Node {

    int v, mx, mn, id, dp;

}a[N];

struct BIT {

    int bit[N];

    void clr(int x) {

        while(x < N) bit[x] = , x += x&-x;

    }

    void add(int x, int a) {

        while(x < N) bit[x] = max(bit[x], a), x += x&-x;

    }

    int mx(int x) {

        int res = ;

        while(x) res = max(res, bit[x]), x -= x&-x;

        return res;

    }

}b;

bool cmp1(Node a, Node b) {

    return a.mx < b.mx;

}

bool cmp2(Node a, Node b) {

    return a.v < b.v;

}

bool cmp3(Node a, Node b) {

    return a.id < b.id;

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m);

    sort(a+l, a+m+, cmp1);

    sort(a+m+, a+r+, cmp2);

    int p = l, q = m+;

    while(q <= r) {

        while(p <= m && a[p].mx <= a[q].v) b.add(a[p].v, a[p].dp), ++p;

        a[q].dp = max(a[q].dp, b.mx(a[q].mn)+);

        ++q;

    }

    for (int i = l; i < p; ++i) b.clr(a[i].v);

    sort(a+m+, a+r+, cmp3);

    cdq(m+, r);

}

int n, m, x, y;

int main() {

    scanf("%d %d", &n, &m);

    for (int i = ; i <= n; ++i) {

        scanf("%d", &a[i].v);

        a[i].mn = a[i].mx = a[i].v;

        a[i].id = i;

        a[i].dp = ;

    }

    for (int i = ; i <= m; ++i) {

        scanf("%d %d", &x, &y);

        a[x].mx = max(a[x].mx, y);

        a[x].mn = min(a[x].mn, y);

    }

    cdq(, n);

    int ans = ;

    for (int i = ; i <= n; ++i) ans = max(ans, a[i].dp);

    printf("%d\n", ans);

    return ;

}

例题2：HDU - 4742

思路：同样是dp，用bit维护最大值和最大值的个树，有点小技巧

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + ;

struct Node {

    int x, y, z, len, cnt;

}a[N];

vector<int> vc;

int T, n, sz;

struct BIT {

    pii bit[N];

    void modify(pii &a, pii b) {

        if(b.fi == a.fi) a.se += b.se;

        else if(b.fi >  a.fi) a = b;

    }

    void clr(int x) {

        while(x <= sz) bit[x] = {, }, x += x&-x;

    }

    void add(int x, pii a) {

        while( x <= sz) modify(bit[x], a), x += x&-x;

    }

    pii mx(int x) {

        pii res = {, };

        while(x) modify(res, bit[x]), x -= x&-x;

        return res;

    }

}b;

bool cmp1(Node a, Node b) {

    return a.x < b.x;

}

bool cmp2(Node a, Node b) {

    return a.y < b.y;

}

pii MX(pii a, pii b) {

    if(a.fi == b.fi) return {a.fi, a.se+b.se};

    else if(a.fi > b.fi) return a;

    else return b;

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(l, m);

    sort(a+l, a+m+, cmp2);

    sort(a+m+, a+r+, cmp2);

    int p = l, q = m+;

    while(q <= r) {

        while(p <= m && a[p].y <= a[q].y) b.add(a[p].z, {a[p].len, a[p].cnt}), ++p;

        pii t1 = b.mx(a[q].z);

        t1.fi++;

        pii t2 = MX({a[q].len, a[q].cnt}, t1);

        a[q].len = t2.fi, a[q].cnt = t2.se;

        ++q;

    }

    for (int i = l; i < p; ++i) b.clr(a[i].z);

    sort(a+m+, a+r+, cmp1);

    cdq(m+, r);

}

int main() {

    scanf("%d", &T);

    while(T--) {

        scanf("%d", &n);

        vc.clear();

        for (int i = ; i <= n; ++i) {

            scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].z);

            vc.pb(a[i].z);

            a[i].len = a[i].cnt = ;

        }

        sort(vc.begin(), vc.end());

        vc.erase(unique(vc.begin(), vc.end()), vc.end());

        for (int i = ; i <= n; ++i) a[i].z = lower_bound(vc.begin(), vc.end(), a[i].z) - vc.begin() + ;

        sz = (int)vc.size();

        sort(a+, a++n, cmp1);

        cdq(, n);

        pii ans = {, };

        for (int i = ; i <= n; ++i) ans = MX(ans, {a[i].len, a[i].cnt});

        printf("%d %d\n", ans.fi, ans.se);

    }

    return ;

}

例题3：HDU - 5324

思路：由于要求字典序最小，所以要求dp[i]：以i为起点的最长上升序列，cdq的话先求右边再求左边

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 5e4 + ;

struct Node {

    int L, R, id;

}a[N];

vector<int> vc, res;

int sz, n, dp[N], pre[N];

struct BIT {

    pii bit[N];

    void clr(int x) {

        while(x <= sz) bit[x] = {, }, x += x&-x;

    }

    void add(int x, pii a) {

        while(x <= sz) {

            if(a.fi > bit[x].fi) bit[x] = a;

            else if(a.fi == bit[x].fi && a.se < bit[x].se) bit[x] = a;

            x += x&-x;

        }

    }

    pii mx(int x) {

        pii res = {, };

        while(x) {

            if(bit[x].fi > res.fi) res = bit[x];

            else if(bit[x].fi == res.fi && bit[x].se < res.se) res = bit[x];

            x -= x&-x;

        }

        return res;

    }

}b;

bool cmp1(const Node &a, const Node &b) {

    return a.id < b.id;

}

bool cmp2(const Node &a, const Node &b) {

    return a.R > b.R;

}

void cdq(int l, int r) {

    if(l == r) return ;

    int m = l+r >> ;

    cdq(m+, r);

    sort(a+l, a+m+, cmp2);

    sort(a+m+, a+r+, cmp2);

    int p = l, q = m+;

    while(p <= m) {

        while(q <= r && a[q].R >= a[p].R) b.add(a[q].L, {dp[a[q].id], a[q].id}), ++q;

        pii P = b.mx(a[p].L);

        if(P.fi+ > dp[a[p].id]) {

            dp[a[p].id] = P.fi+;

            pre[a[p].id] = P.se;

        }

        else if(P.fi+ == dp[a[p].id] && P.se < pre[a[p].id]) pre[a[p].id] = P.se;

        ++p;

    }

    for (int i = m+; i < q; ++i) b.clr(a[i].L);

    sort(a+l, a+m+, cmp1);

    cdq(l, m);

}

int main() {

    while(~scanf("%d", &n)) {

        vc.clear();

        for (int i = ; i <= n; ++i) scanf("%d", &a[i].L), vc.pb(a[i].L);

        for (int i = ; i <= n; ++i) scanf("%d", &a[i].R);

        sort(vc.begin(), vc.end());

        vc.erase(unique(vc.begin(), vc.end()), vc.end());

        sz = (int)vc.size();

        for (int i = ; i <= n; ++i) {

            a[i].id = i;

            dp[i] = ;

            pre[i] = ;

            a[i].L = lower_bound(vc.begin(), vc.end(), a[i].L) - vc.begin() + ;

        }

        cdq(, n);

        res.clear();

        int mx = , pe = ;

        res.pb();

        for (int i = ; i <= n; ++i) {

            if(dp[i] > mx) {

                mx = dp[i];

                pe = pre[i];

                res[] = i;

            }

        }

        printf("%d\n", mx);

        while(pe) {

            res.pb(pe);

            pe = pre[pe];

        }

        for (int i = ; i < mx; ++i) printf("%d%c", res[i], " \n"[i==mx-]);

    }

    return ;

}

参考：https://www.cnblogs.com/Sakits/p/7990875.html

整体二分：

1.静态区间第k大

思路：首先对于单个查询我们可以二分答案求解，对于一个二分出来的mid，如果我们能知道小于等于mid的个数，就能判断答案是小于等于mid还是大于mid

整体二分就是将所有查询的二分放在一起考虑，如果对于某个查询我们能快速知道小于等于mid的个数（可以用树状数组维护，将小于等于mid的修改按位置加入），

我们就能知道哪些查询答案是小于等于mid，哪些查询答案是大于mid，哪些修改只对答案小于等于mid的查询有用，哪些修改只对答案大于mid的查询有用

按照这个标准将所有查询修改分成两部分，递归求解，直到二分答案的区间长度为1。具体实现看代码（l,r表示二分答案的区间，L,R表示查询或修改的区间）

时间复杂度：与cdq分治类似，为O(n*log(n)^2)

POJ - 2104

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1.1e5 + ;

const int INF = 0x3f3f3f3f;

struct Node {

    int ty, x, y, k, id;

}a[N], t1[N], t2[N];

int n, m, ans[N];

struct BIT {

    int bit[N];

    inline void add(int x, int a) {

        while(x < N) bit[x] += a, x += x&-x;

    }

    inline int sum(int x) {

        int res = ;

        while(x) res += bit[x], x -= x&-x;

        return res;

    }

}b;

void solve(int l, int r, int L, int R) {

    if(l > r || L > R) return ;

    if(l == r) {

        for (int i = L; i <= R; ++i) if(a[i].ty == ) ans[a[i].id] = l;

        return ;

    }

    int m = l+r >> ;

    int p = , q = ;

    for (int i = L; i <= R; ++i) {

        if(a[i].ty == ){

            int cnt = b.sum(a[i].y) - b.sum(a[i].x-);

            if(cnt >= a[i].k) t1[++p] = a[i];

            else a[i].k -= cnt, t2[++q] = a[i];

        }

        else {

            if(a[i].x <= m) b.add(a[i].id, ), t1[++p] = a[i];

            else t2[++q] = a[i];

        }

    }

    for (int i = ; i <= p; ++i) if(t1[i].ty == ) b.add(t1[i].id, -);

    for (int i = ; i <= p; ++i) a[L+i-] = t1[i];

    for (int i = ; i <= q; ++i) a[L+p+i-] = t2[i];

    solve(l, m, L, L+p-);

    solve(m+, r, L+p, R);

}

int main() {

    scanf("%d %d", &n, &m);

    for (int i = ; i <= n; ++i) {

        scanf("%d", &a[i].x);

        a[i].ty = ;

        a[i].id = i;

    }

    for (int i = n+; i <= n+m; ++i) {

        scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].k);

        a[i].ty = ;

        a[i].id = i-n;

    }

    solve(-INF, INF, , n+m);

    for (int i = ; i <= m; ++i) printf("%d\n", ans[i]);

    return ;

}

2.动态第k大

思路：在静态第k大的基础上将修改拆成两个，具体看代码

ZOJ - 2112

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + ;

const int INF = 0x3f3f3f3f;

struct Node {

    int ty, x, y, k, id;

}q[N], t1[N], t2[N];

int n, m, a[N], ans[N];

struct BIT {

    int bit[N];

    inline void add(int x, int a) {

        while(x <= n) bit[x] += a, x += x&-x;

    }

    inline int sum(int x) {

        int res = ;

        while(x) res += bit[x], x -= x&-x;

        return res;

    }

}b;

void solve(int l, int r, int L, int R) {

    if(l > r || L > R) return ;

    if(l == r) {

        for (int i = L; i <= R; ++i) if(q[i].ty == ) ans[q[i].id] = l;

        return ;

    }

    int m = l+r >> , c1 = , c2 = ;

    for (int i = L; i <= R; ++i) {

        if(q[i].ty == ) {

            int cnt = b.sum(q[i].y) - b.sum(q[i].x-);

            if(q[i].k <= cnt) t1[++c1] = q[i];

            else q[i].k -= cnt, t2[++c2] = q[i];

        }

        else {

            if(q[i].x <= m) b.add(q[i].id, q[i].y), t1[++c1] = q[i];

            else t2[++c2] = q[i];

        }

    }

    for (int i = ; i <= c1; ++i) if(t1[i].ty == ) b.add(t1[i].id, -t1[i].y);

    for (int i = ; i <= c1; ++i) q[L+i-] = t1[i];

    for (int i = ; i <= c2; ++i) q[L+c1+i-] = t2[i];

    solve(l, m, L, L+c1-);

    solve(m+, r, L+c1, R);

}

int T, x, y, k;

char op[];

int main() {

    scanf("%d", &T);

    while(T--) {

        scanf("%d %d", &n, &m);

        int tot = ;

        for (int i = ; i <= n; ++i) {

            scanf("%d", &a[i]);

            q[++tot] = {, a[i], , , i};

        }

        for (int i = ; i <= m; ++i) {

            scanf("%s", op);

            if(op[] == 'Q') {

                scanf("%d %d %d", &x, &y, &k);

                q[++tot] = {, x, y, k, i};

            }

            else {

                scanf("%d %d", &x, &y);

                q[++tot] = {, a[x], -, , x};

                q[++tot] = {, a[x]=y, , , x};

            }

            ans[i] = -;

        }

        solve(, INF, , tot);

        for (int i = ; i <= m; ++i) if(~ans[i]) printf("%d\n", ans[i]);

    }

    return ;

}

例题1：HYSBZ - 3110

思路：在线段树上进行区间加法

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

//#define mp make_pair

#define pb push_back

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdi pair<double, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 5e4 + ;

struct Node {

    int ty, x, y, id;

    LL c;

}a[N], t1[N], t2[N];

int n, m, ans[N];

LL tree[N<<], lazy[N<<];

inline void push_up(int rt) {

    tree[rt] = tree[rt<<] + tree[rt<<|];

}

inline void push_down(int rt, int len) {

    lazy[rt<<] += lazy[rt];

    lazy[rt<<|] += lazy[rt];

    tree[rt<<] += lazy[rt]*(len - (len>>));

    tree[rt<<|] += lazy[rt]*(len>>);

    lazy[rt] = ;

}

void update(int L, int R, int x, int rt, int l, int r) {

    if(L <= l && r <= R) {

        lazy[rt] += x;

        tree[rt] += x*(r-l+);

        return ;

    }

    if(lazy[rt]) push_down(rt, r-l+);

    int m = l+r >> ;

    if(L <= m) update(L, R, x, ls);

    if(R > m) update(L, R, x, rs);

    push_up(rt);

}

LL query(int L, int R, int rt, int l, int r) {

    if(L <= l && r <= R) return tree[rt];

    if(lazy[rt]) push_down(rt, r-l+);

    int m = l+r >> ;

    LL ans = ;

    if(L <= m) ans += query(L, R, ls);

    if(R > m) ans += query(L, R, rs);

    push_up(rt);

    return ans;

}

void solve(int l, int r, int L, int R) {

    if(l > r || L > R) return ;

    if(l == r) {

        for (int i = L; i <= R; ++i) if(a[i].ty == ) ans[a[i].id] = l;

        return ;

    }

    int m = l+r >> , p = , q = ;

    for (int i = L; i <= R; ++i) {

        if(a[i].ty == ) {

            LL cnt = query(a[i].x, a[i].y, , , n);

            if(cnt >= a[i].c) t2[++q] = a[i];

            else a[i].c -= cnt, t1[++p] = a[i];

        }

        else {

            if(a[i].c > m) update(a[i].x, a[i].y, , , , n), t2[++q] = a[i];

            else t1[++p] = a[i];

        }

    }

    for (int i = ; i <= q; ++i) if(t2[i].ty == ) update(t2[i].x, t2[i].y, -, , , n);

    for (int i = ; i <= p; ++i) a[L+i-] = t1[i];

    for (int i = ; i <= q; ++i) a[L+p+i-] = t2[i];

    solve(l, m, L, L+p-);

    solve(m+, r, L+p, R);

}

int tot = ;

int main() {

    scanf("%d %d", &n, &m);

    for (int i = ; i <= m; ++i) scanf("%d %d %d %lld", &a[i].ty, &a[i].x, &a[i].y, &a[i].c), a[i].id = (a[i].ty == )?i:(++tot);

    solve(-n, n, , m);

    for (int i = ; i <= tot; ++i) printf("%d\n", ans[i]);

    return ;

}

例题2：P1527 [国家集训队]矩阵乘法

思路：在二维树状数组上进行加法

代码：

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define LL long long 

const int N = , M = 3.2e5 + ;

const int INF = 0x3f3f3f3f;

struct Node {

    int ty, x1, y1, x2, y2, k, id;

}a[M], t1[M], t2[M];

int n, m, ans[M];

struct BIT2 {

    int bit[N][N];

    inline void add(int x, int y, int a) {

        for (int i = x; i <= n; i += i&-i) {

            for (int j = y; j <= n; j += j&-j) {

                bit[i][j] += a;

            }

        }

    }

    inline int sum(int x, int y) {

        int res = ;

        for (int i = x; i > ; i -= i&-i) {

            for (int j = y; j > ; j -= j&-j) {

                res += bit[i][j];

            }

        }

        return res;

    }

}B;

void solve(int l, int r, int L, int R){

    if(l > r || L > R) return ;

    if(l == r) {

        for (int i = L; i <= R; ++i) {

            if(a[i].ty == ) ans[a[i].id] = l;

        }

        return ;

    }

    int m = l+r >> , p = , q = ;

    for (int i = L; i <= R; ++i) {

        if(a[i].ty == ) {

            if(a[i].k <= m) B.add(a[i].x1, a[i].y1, ), t1[++p] = a[i];

            else t2[++q] = a[i];

        }

        else {

            int cnt = B.sum(a[i].x2, a[i].y2) - B.sum(a[i].x1-, a[i].y2) - B.sum(a[i].x2, a[i].y1-) + B.sum(a[i].x1-, a[i].y1-);

            if(cnt >= a[i].k) t1[++p] = a[i];

            else a[i].k -= cnt, t2[++q] = a[i];

        }

    }

    for (int i = ; i <= p; ++i) if(t1[i].ty == ) B.add(t1[i].x1, t1[i].y1, -);

    for (int i = ; i <= p; ++i) a[L+i-] = t1[i];

    for (int i = ; i <= q; ++i) a[L+p+i-] = t2[i];

    solve(l, m, L, L+p-);

    solve(m+, r, L+p, R);

}

int main() {

    int tot = ;

    scanf("%d %d", &n, &m);

    for (int i = ; i <= n; ++i) {

        for (int j = ; j <= n; ++j) {

            ++tot;

            scanf("%d", &a[tot].k);

            a[tot].x1 = i;

            a[tot].y1 = j;

            a[tot].ty = ;

        }

    }

    for (int i = ; i <= m; ++i) {

        ++tot;

        scanf("%d %d %d %d %d", &a[tot].x1, &a[tot].y1, &a[tot].x2, &a[tot].y2, &a[tot].k);

        a[tot].ty = ;

        a[tot].id = i;

    }

    solve(-INF, INF, , tot);

    for (int i = ; i <= m; ++i) printf("%d\n", ans[i]);

    return ;

}

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