BZOJ1452——[JSOI2009]Count

1、题目大意： 就是给一个n×m的方格，然后一些平面上的 求和 修改操作

2、分析：二维树状数组裸题

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int C[110][310][310];
int a[310][310];
int n, m;
inline void add(int c, int x, int y, int z){
    for(; x <= n; x += (x & -x)){
        int yy = y;
        for(; yy <= m; yy += (yy & -yy)){
            C[c][x][yy] += z;
        }
    }
    return;
}
inline int query(int c, int x, int y){
    int res = 0;
    for(; x > 0; x -= (x & -x)){
        int yy = y;
        for(; yy > 0; yy -= (yy & -yy)){
            res += C[c][x][yy];
        }
    }
    return res;
}
int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            int c; scanf("%d", &c);
            add(c, i, j, 1); a[i][j] = c;
        }
    }
    int Q;
    scanf("%d", &Q);
    while(Q --){
        int op, x1, y1, x2, y2, c;
        scanf("%d", &op);
        if(op == 1){
            scanf("%d%d%d", &x1, &y1, &c);
            add(a[x1][y1], x1, y1, -1);
            add(c, x1, y1, 1);
            a[x1][y1] = c;
        }
        else{
            scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &c);
            printf("%d\n", query(c, x2, y2) - query(c, x1 - 1, y2) - query(c, x2, y1 - 1) + query(c, x1 - 1, y1 - 1));
        }
    }
    return 0;
}