BZOJ1176 [Balkan2007]Mokia 【CDQ分治】

题目

维护一个W*W的矩阵，初始值均为S.每次操作可以增加某格子的权值,或询问某子矩阵的总权值.修改操作数M<=160000,询问数Q<=10000,W<=2000000.

输入格式

第一行两个整数,S,W;其中S为矩阵初始值;W为矩阵大小

接下来每行为一下三种输入之一(不包含引号):

“1 x y a”

“2 x1 y1 x2 y2”

“3”

输入1:你需要把(x,y)(第x行第y列)的格子权值增加a

输入2:你需要求出以左下角为(x1,y1),右上角为(x2,y2)的矩阵内所有格子的权值和,并输出

输入3:表示输入结束

输出格式

对于每个输入2,输出一行,即输入2的答案

输入样例

0 4

1 2 3 3

2 1 1 3 3

1 2 2 2

2 2 2 3 4

3

输出样例

3

5

提示

保证答案不会超过int范围

题解

CDQ分治
每个询问分解成4个前缀和
然后每个询问或修改就有三维(x,y,t)
一个询问被一个修改影响当且仅当t1 > t2 且x1 >= x2 且 y1 >= y2
可以上CDQ分治
【可以顺便吧2683A了】

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define lbt(x) (x & -x)
using namespace std;
const int maxn = 200005,maxm = 2000005,INF = 1000000000;
inline int RD(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}
return out * flag;
}
LL S[maxm],N,s;
void add(int u,int v){while (u <= N) S[u] += v,u += lbt(u);}
LL sum(int u){LL ans = 0; while (u) ans += S[u],u -= lbt(u); return ans;}

struct Que{LL x,y,id,v,pos,t;}Q[maxn],T[maxn];
inline bool operator <(const Que& a,const Que& b){
if (a.x == b.x && a.y == b.y) return a.t < b.t;
if (a.x == b.x) return a.y < b.y;
return a.x < b.x;
}
LL Qi = 0,ans[maxn];
void CDQ(int l,int r){
if (l == r) return;
int mid = l + r >> 1,l1 = l,l2 = mid + 1;
for (int i = l; i <= r; i++)
if (Q[i].t <= mid && !Q[i].id) add(Q[i].y,Q[i].v);
else if (Q[i].t > mid && Q[i].id) ans[Q[i].id] += Q[i].pos * sum(Q[i].y);
for (int i = l; i <= r; i++){
//printf("[%lld,%lld],id = %lld,t = %lld\n",Q[i].x,Q[i].y,Q[i].id,Q[i].t);
if (Q[i].t <= mid){
T[l1++] = Q[i];
if (!Q[i].id) add(Q[i].y,-Q[i].v);
}
else T[l2++] = Q[i];
}
for (int i = l; i <= r; i++) Q[i] = T[i];
CDQ(l,mid); CDQ(mid + 1,r);
}
int main(){
//freopen("in.txt","r",stdin);
//freopen("out1.txt","w",stdout);
s = RD(); N = RD();
LL opt,a,b,c,d;
while (true){
opt = RD(); if (opt == 3) break;
if (opt & 1){
a = RD(),b = RD(),c = RD();
Q[++Qi] = (Que){a,b,0,c,0,Qi};
}else {
a = RD(),b = RD(),c = RD(); d = RD();
int pos = ++ans[0];
Q[++Qi] = (Que){c,d,pos,0,1,Qi};
Q[++Qi] = (Que){c,b - 1,pos,0,-1,Qi};
Q[++Qi] = (Que){a - 1,d,pos,0,-1,Qi};
Q[++Qi] = (Que){a - 1,b - 1,pos,0,1,Qi};
ans[pos] = (c - a + 1) * (d - b + 1) * s;
}
}
sort(Q + 1,Q + 1 + Qi);
CDQ(1,Qi);
REP(i,ans[0]) printf("%lld\n",ans[i]);
return 0;
}