luogu P4390 [BOI2007]Mokia 摩基亚

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昨天做完三维偏序并不能理解CDQ

今天做了这个题才行

(觉得没理解三维偏序是因为二维偏序没按正常方式理解)

CDQ分治应用于数据结构 适用于离线的题

原理是后面的询问只能被前面的修改影响

可以分治处理 分治的左区间的修改会影响右区间的询问

这样就可以优化掉一维树状数组

比如说这个题 二维树状数组开不下

CDQ分治秒

我们知道二维区间和可以转成前缀和

然后只有x,y,时间都比查询小的修改才会算进去

所以树状数组维护正序对就OK啦(三维偏序)

我是按x二分 因为时间周自然排好序

Code:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<ctime>
#include<algorithm>
#include<iostream>
#include <bits/stdc++.h>
#define D double
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 1e8
using namespace std;
typedef long long ll;
#define Br puts("GG")
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 200005;
//head
int n,maxm,T;
struct node {
    int x,y,c,op,idx;
} a[N];
node tmp[N];
int t[N<<4],ans[N];
void upd(int x,int k) {
    for(int i = x;i <= maxm;i += (i & (-i))) t[i] += k;
}
int qry(int x) {
    int sum = 0;
    for(int i = x;i;i -= (i & (-i))) sum += t[i];
    return sum;
}

void CDQ(int l,int r) {
    if(l == r) return;
    int m = l+r >> 1;
    CDQ(l,m),CDQ(m+1,r);
    int p1 = l,p2 = m+1;
    rep(i,l,r) {
        if(p1 <= m && (p2 > r || a[p1].x <= a[p2].x)) {
            if(!a[p1].op) upd(a[p1].y,a[p1].c);
            tmp[i] = a[p1++];
        } else {
            if(a[p2].op) ans[a[p2].idx] += a[p2].c * qry(a[p2].y);
            tmp[i] = a[p2++];
        }
    }
    rep(i,l,m) if(a[i].op == 0) upd(a[i].y,-a[i].c);
    rep(i,l,r) a[i] = tmp[i];
}

int main() {
    int op;
    n = read(),maxm = read();
    while((op = read()) ^ 3) {
        if(op == 1) {
            a[++n].x = read();a[n].y = read();
            a[n].c = read();
            a[n].op = a[n].idx = 0;
        } else {
            T++;
            int x = read(),y = read(),A = read(),b = read();
            a[++n] = (node){A,b,1,1,T};
            a[++n] = (node){A,y - 1,-1,1,T};
            a[++n] = (node){x - 1,b,-1,1,T};
            a[++n] = (node){x - 1,y - 1,1,1,T};
        }
    }
    CDQ(1,n);
    rep(i,1,T) printf("%d
",ans[i]);
    return 0;
}