BZOJ 1176: [Balkan2007]Mokia CDQ分治，容斥

Description

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.
Input

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
Output

对于每个输入2,输出一行,即输入2的答案
Sample Input
0 4

1 2 3 3

2 1 1 3 3

1 2 2 2

2 2 2 3 4

3
Sample Output
3

5

解题方法： 给个别人的链接
初学CDQ花了快1天理解这个原理，再推荐一位大牛写的关于CDQ分治的总结: 见这里 初级CDQ大概理解了，不过CDQ套CDQ，以及高维偏序还不理解，强撸题吧！注意这个题数组开小不是RE,是WA，神坑。

//bzoj 1176//1维排序，二维分治，3维BIT#include <bits/stdc++.h>using namespace std;const int maxq = 200010;const int maxn = 500010;int S, W, ans[maxq];inline int read(){    char ch=getchar();    int f=1,x=0;    while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');ch=getchar();}    return x*f;}namespace BIT{    int tree[2000010];    inline int lowbit(int x) {return x & -x;}    inline void update(int x, int y) {for(int i = x; i <= W; i+=lowbit(i)) tree[i] += y;}    inline int query(int x) {int res = 0; for(int i = x; i; i -= lowbit(i)) res += tree[i]; return res;}}using namespace BIT;struct Q{    int id, x, y, opt, del, ID;    Q(){}    Q(int id, int x, int y, int opt, int del, int ID) : id(id), x(x), y(y), opt(opt), del(del), ID(ID) {}    bool operator < (const Q &rhs) const{        return (x == rhs.x && y == rhs.y) ? opt < rhs.opt : ((x == rhs.x) ? y < rhs.y : x < rhs.x);    }}q[maxq*4], tmp[maxq*4];void CDQ(int l, int r){    if(l == r) return ;    int mid = (l+r)>>1, z1 = l, z2 = mid + 1;    for(int i = l; i <= r; i++){        if(q[i].opt == 1 && q[i].id <= mid) update(q[i].y, q[i].del);        if(q[i].opt == 2 && q[i].id > mid) ans[q[i].ID] += query(q[i].y) * q[i].del;    }    for(int i = l; i <= r; i++) if(q[i].opt == 1 && q[i].id <= mid) update(q[i].y, -q[i].del);    for(int i = l; i <= r; i++) if(q[i].id <= mid) tmp[z1++] = q[i]; else tmp[z2++] = q[i];    for(int i = l; i <= r; i++) q[i] = tmp[i];    CDQ(l, mid);    CDQ(mid + 1, r);}int t, z;void presolve(int x1, int y1, int x2, int y2){ //操作拆分 sum(x2,y2)+sum(x1-1,y1-1)+sum(x1-1,y2)-sum(x2,y1-1)    z++;    t++; q[t].id = t; q[t].ID = z; q[t].x = x1 - 1; q[t].y = y1 - 1; q[t].del = 1; q[t].opt = 2;    t++; q[t].id = t; q[t].ID = z; q[t].x = x2; q[t].y = y2; q[t].del = 1; q[t].opt = 2;    t++; q[t].id = t; q[t].ID = z; q[t].x = x1 - 1; q[t].y = y2; q[t].del = -1; q[t].opt = 2;    t++; q[t].id = t; q[t].ID = z; q[t].x = x2; q[t].y = y1 - 1; q[t].del = -1; q[t].opt = 2;}int main(){    //scanf("%d%d", &S, &W);    S = read(); W = read();    t = z = 0;    while(1){        int cmd;        //scanf("%d", &cmd);        cmd = read();        if(cmd == 3) break;        if(cmd == 1){t++; //scanf("%d%d%d", &q[t].x, &q[t].y, &q[t].del);        q[t].x = read(), q[t].y = read(), q[t].del = read();        q[t].id = t; q[t].opt = 1; }        if(cmd == 2){            int x1, x2, y1, y2;            //scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            x1 = read(), y1 = read(), x2 = read(), y2 = read();            presolve(x1, y1, x2, y2);        }    }    sort(q + 1, q + t + 1);    CDQ(1, t);    for(int i = 1; i <= z; i++) printf("%d\n", ans[i]);    return 0;}