bzoj3110: [Zjoi2013]K大数查询 【cdq分治&树套树】

　　模板题，折腾了许久。

　　cqd分治整体二分，感觉像是把询问分到答案上。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 template <typename T> void Max(T &a, T &b) { if (a < b) a = b; }
 template <typename T> void Min(T &a, T &b) { if (a > b) a = b; }
 //******************************************

 const int maxn = ;

 struct Seg_Tree {
     int sum[maxn << ], lazy[maxn << ];
     void Push_down(int o, int m) {
         if (!lazy[o]) return;
         lazy[o << ] += lazy[o], lazy[o <<  | ] += lazy[o];
         sum[o << ] += lazy[o] * (m - (m >> )), sum[o <<  | ] += lazy[o] * (m >> );
         lazy[o] = ;
     }
     void Push_up(int o) { sum[o] = sum[o << ] + sum[o <<  | ]; }
     void update(int o, int l, int r, int ql, int qr, int v) {
         if (ql <= l && r <= qr) {
             lazy[o] += v;
             sum[o] += v * (r - l + );
             return;
         }
         Push_down(o, r - l + );
         int mid = l + r >> ;
         if (ql <= mid) update(o << , l, mid, ql, qr, v);
         if (qr > mid) update(o <<  | , mid + , r, ql, qr, v);
         Push_up(o);
     }
     int query(int o, int l, int r, int ql, int qr) {
         if (ql <= l && r <= qr) return sum[o];
         Push_down(o, r - l + );
         int mid = l + r >> ;
         int ret();
         if (ql <= mid) ret += query(o << , l, mid, ql, qr);
         if (qr > mid) ret += query(o <<  | , mid + , r, ql, qr);
         return ret;
     }
     void CLR() { clr(sum), clr(lazy); }
 } T;

 struct Complex {
     int flag, x, y, c, id;
     inline bool operator < (const Complex &A) const {
         return id < A.id;
     }
 } src[maxn];

 int N, ans[maxn], v[maxn];
 void cdq(int ansl, int ansr, int l, int r) {
     if (ansl == ansr) {
         rep(i, l, r) if (src[i].flag == ) ans[src[i].id] = ansl;
         return;
     }
     if (l > r) return;
     sort(src + l, src + r + );
     int m = ansl + ansr +  >> ;
     int j = l;
     rep(i, l, r) {
         if (src[i].flag == ) {
             if (src[i].c >= m) T.update(, , N, src[i].x, src[i].y, ), v[i] = ;
             else v[i] = ;
         }
         else {
             int t = T.query(, , N, src[i].x, src[i].y);
             if (t >= src[i].c) v[i] = ;
             else src[i].c -= t, v[i] = ;
         }
         j += !v[i];
     }
     rep(i, l, r)
         if (src[i].flag == )
             if (src[i].c >= m) T.update(, , N, src[i].x, src[i].y, -);
     int t = j - ;
     if (j != l)
     rep(i, l, t) {
         while (j < r && v[j]) ++j;
         if (v[i]) swap(src[i], src[j]), swap(v[i], v[j]), ++j;
     }
     cdq(ansl, m - , l, t);
     cdq(m, ansr, t + , r);
 }

 int read() {
     int l = , s = ; char ch = getchar();
     while (ch < '' || ch > '') { if (ch == '-') l = -; ch = getchar(); }
     while (ch >= '' && ch <= '') { s = (s << ) + (s << ) + ch - ''; ch = getchar(); }
     return s * l;
 }
 int main() {
     int n, m; scanf("%d%d", &n, &m);
     rep(i, , m) scanf("%d%d%d%d", &src[i].flag, &src[i].x, &src[i].y, &src[i].c), src[i].id = i;
     N = n;
     cdq(-n, n, , m);
     sort(src + , src +  + m);
     rep(i, , m) if (src[i].flag == ) printf("%d\n", ans[i]);
     return ;
 }

　　树套树本来想把代表区间的那个线段树开在外面，不是不可以，但是lazy标记可能要一个vector才能存，因为我要存多个种类。

　　然后就只能把代表数字那一维开在外面，代表这一段数字在1-n这个区间的分布情况，这样就可以在第二层线段树上打lazy标记。

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (register int i = a; i <= b; i++)
 #define drep(i, a, b) for (register int i = a; i >= b; i--)
 #define REP(i, a, b) for (register int i = a; i < b; i++)
 #define pb push_back
 #define mp make_pair
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 const int inf = ~0U >> ;
 const i64 INF = ~0ULL >> ;
 //*******************************

 const int maxn = , maxnn = ;

 int root[maxn << ];
 int ls[maxnn], rs[maxnn], sum[maxnn], lazy[maxnn];

 int ndtot, n, N;
 inline void Push_up(int o) { sum[o] = sum[ls[o]] + sum[rs[o]]; }
 inline void Push_down(int o, int m) {
     if (!lazy[o]) return;
     if (!ls[o]) ls[o] = ++ndtot;
     if (!rs[o]) rs[o] = ++ndtot;
     lazy[ls[o]] += lazy[o], lazy[rs[o]] += lazy[o];
     sum[ls[o]] += lazy[o] * (m - (m >> )), sum[rs[o]] += lazy[o] * (m >> );
     lazy[o] = ;
 }
 void update(int &k, int l, int r, int ql, int qr, int v) {
     if (!k) k = ++ndtot;
     if (ql <= l && r <= qr) {
         sum[k] += v * (r - l + );
         lazy[k] += v;
         return;
     }
     int mid = l + r >> ;
     Push_down(k, r - l + );
     if (ql <= mid) update(ls[k], l, mid, ql, qr, v);
     if (qr > mid) update(rs[k], mid + , r, ql, qr, v);
     Push_up(k);
 }
 void insrt(int o, int l, int r, int ql, int qr, int c) {
     while (l != r) {
         int mid = l + r >> ;
         update(root[o], , n, ql, qr, );
         if (c <= mid) o <<= , r = mid;
         else o = o <<  | , l = mid + ;
     }
     update(root[o], , n, ql, qr, );
 }

 int query(int o, int l, int r, int ql, int qr) {
     if (!o) return ;
     if (ql <= l && r <= qr) return sum[o];
     Push_down(o, r - l + );
     int mid = l + r >> ;
     int ret();
     if (ql <= mid) ret += query(ls[o], l, mid, ql, qr);
     if (qr > mid) ret += query(rs[o], mid + , r, ql, qr);
     return ret;
 }
 int solve(int o, int l, int r, int ql, int qr, int c) {
     while (l != r) {
         int mid = l + r >> ;
         int t = query(root[o <<  | ], , n, ql, qr);
         if (t >= c) l = mid + , o = o <<  |;
         else r = mid, o = o << , c -= t;
     }
     return l;
 }

 int main() {
     int m; scanf("%d%d", &n, &m);
     N =  * n + ;
     while (m--) {
         int flag, a, b, c; scanf("%d%d%d%d", &flag, &a, &b, &c);
         if (flag == ) {
             c += n + ;
             insrt(, , N, a, b, c);
         }
         else printf("%d\n", solve(, , N, a, b, c) - n - );
     }
 }