[BZOJ 3262]陌上开花

今天写了到偏序问题，发现博主真的是个傻X

传送门

以前的写法

/**************************************************************
    Problem: 3262
    User: MiEcoku
    Language: C++
    Result: Accepted
    Time:3416 ms
    Memory:6296 kb
****************************************************************/

#include <cstdio>
#include <algorithm>
using namespace std;
#define mid ( l + r >> 1)
#define lowbit(x) ( x & -x)

struct GG {
    int x, y, z, id, w;
    void get () {
        scanf("%d%d%d", &x, &y, &z);
    }
}a[], b[];

], k;
; ; x -= lowbit(x)) ans += dis[x]; return ans; }
void add(int x, int y) { for ( ; x <= k; x += lowbit(x)) dis[x] += y; }

bool cmp1(GG a, GG b) { return a.x < b.x || (a.x == b.x && a.y < b.y) || (a.x == b.x && a.y == b.y && a.z < b.z); }
bool cmp2(GG a, GG b) { return a.y < b.y || (a.y == b.y && a.z < b.z); }

];
void cdq(int l, int r) {
    , r);
    sort(b+l, b+mid+, cmp2); sort(b+mid+, b+r+, cmp2);
    ;
    for ( int i = l; i <= mid; i ++) {
        while( b[q].y < b[i].y && q <= r) ans[b[q].id] += query(b[q].z), q ++;
        add(b[i].z, b[i].w);
    }
    while( q <= r) ans[b[q].id] += query(b[q].z), q ++;
    for ( int i = l; i <= mid; i ++) add(b[i].z, -b[i].w);
}

];
int main() {
    scanf("%d%d", &_, &k);
    ; i <= _; i ++) a[i].get();
    sort( a+, a++_, cmp1);
    ; i <= _; i ++) {
        cnt ++;
        ].x || a[i].y != a[i+].y || a[i].z != a[i+].z)
        b[++ n] = a[i], b[n].w = cnt, cnt = , b[n].id = n;
    }
    cdq(, n);
    ; i <= n; i ++)
        d[ans[b[i].id]+b[i].w-] += b[i].w;
    ; i < _; i ++) printf("%d\n", d[i]);
}

可以看出这是个n * logn^2 的算法，这也是博主在大多数地方看到的写法，结果傻叉的以为cdq就是n * long n ^ 2 的算法

实际上可不可以更快？

void cdq(int l, int r) {
    , r);
    sort(b+l, b+mid+, cmp2); sort(b+mid+, b+r+, cmp2);
    ;
    for ( int i = l; i <= mid; i ++) {
        while( b[q].y < b[i].y && q <= r) ans[b[q].id] += query(b[q].z), q ++;
        add(b[i].z, b[i].w);
    }
    while( q <= r) ans[b[q].id] += query(b[q].z), q ++;
    for ( int i = l; i <= mid; i ++) add(b[i].z, -b[i].w);
}

可以看出，在cdq内部排了个序，然而真的有这个必要吗？ 答案是否定的

我们都知道归并排序，可以看出，内部这个排序无非是想要将第二位排序

但在内部这个循环内就已经排序，所以那个排序可以省略

我们新开个数组

void cdq(int l, int r) {
    , r);
    , cnt = ;
    for ( int i = l; i <= mid; i ++) {
        while( b[q].y < b[i].y && q <= r) ans[b[q].id] += query(b[q].z), T[++ cnt] = b[q], q ++;
        add(b[i].z, b[i].w); T[++ cnt] = b[i];
    }
    while( q <= r) ans[b[q].id] += query(b[q].z), T[++ cnt] = b[q], q ++;
    for ( int i = l; i <= mid; i ++) add(b[i].z, -b[i].w);
    cnt = ;
    for ( int i = l; i <= r; ++ i) b[i] = T[++ cnt];
}

这样，我们就省略了内部的一个排序问题，做到n * long n

还能再快不？

能！

    for ( int i = l; i <= mid; i ++) add(b[i].z, -b[i].w);

对于这句话，我们能不能将它优化掉？

我们考虑加入一个时间戳 tim

int query(int x, int K) {
    ;
    ; x -= lowbit(x)) if( mark[x] == K) ans += dis[x];
    return ans;
}
void add(int x, int y, int K) {
    for ( ; x <= k; x += lowbit(x)) if( mark[x] == K) dis[x] += y;
    else mark[x] = K, dis[x] = y;
}

然后对树状数组进行点修改

这样我们就将常数再次优化了下

上总代码

/**************************************************************
    Problem: 3262
    User: MiEcoku
    Language: C++
    Result: Accepted
    Time:1428 ms
    Memory:9028 kb
****************************************************************/

/**************************************************************
    Problem: 3262
    User: MiEcoku
    Language: C++
    Result: Accepted
    Time:3416 ms
    Memory:6296 kb
****************************************************************/

#include <cstdio>
#include <algorithm>
using namespace std;
#define mid ( l + r >> 1)
#define lowbit(x) ( x & -x)

struct GG {
    int x, y, z, id, w;
    void get () {
        scanf("%d%d%d", &x, &y, &z);
    }
}a[], b[], T[];

], k, tim, mark[];
int query(int x, int K) {
    ;
    ; x -= lowbit(x)) if( mark[x] == K) ans += dis[x];
    return ans;
}
void add(int x, int y, int K) {
    for ( ; x <= k; x += lowbit(x)) if( mark[x] == K) dis[x] += y;
    else mark[x] = K, dis[x] = y;
}

bool cmp1(GG a, GG b) { return a.x < b.x || (a.x == b.x && a.y < b.y) || (a.x == b.x && a.y == b.y && a.z < b.z); }
bool cmp2(GG a, GG b) { return a.y < b.y || (a.y == b.y && a.z < b.z); }

];
void cdq(int l, int r) {
    , r);
    , cnt = ; ++ tim;
    for ( int i = l; i <= mid; i ++) {
        while( b[q].y < b[i].y && q <= r) ans[b[q].id] += query(b[q].z, tim), T[++ cnt] = b[q], q ++;
        add(b[i].z, b[i].w, tim); T[++ cnt] = b[i];
    }
    while( q <= r) ans[b[q].id] += query(b[q].z, tim), T[++ cnt] = b[q], q ++;
    cnt = ;
    for ( int i = l; i <= r; ++ i) b[i] = T[++ cnt];
}

];
int main() {
    scanf("%d%d", &_, &k);
    ; i <= _; i ++) a[i].get();
    sort( a+, a++_, cmp1);
    ; i <= _; i ++) {
        cnt ++;
        ].x || a[i].y != a[i+].y || a[i].z != a[i+].z)
        b[++ n] = a[i], b[n].w = cnt, cnt = , b[n].id = n;
    }
    cdq(, n);
    ; i <= n; i ++)
        d[ans[b[i].id]+b[i].w-] += b[i].w;
    ; i < _; i ++) printf("%d\n", d[i]);
}

博主因为傻逼不知道这么写然后考试写的KT-tree被卡了

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