BZOJ 1452 [JSOI2009] Count

题目链接：见这里
解题方法: 二维树状数组，对每个颜色建一个二维树状数组。
代码如下：

#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) x&(-x)
int maze[305][305];
int c[110][305][305];
int n, m, q;

void update(int x, int y, int cc, int val){
    for(int i = x; i <= n; i += lowbit(i)){
        for(int j = y; j <= m; j += lowbit(j)){
            c[cc][i][j] += val;
        }
    }
}

int query(int x, int y, int cc){
    int res = 0;
    for(int i = x; i; i -= lowbit(i)){
        for(int j = y; j; j -= lowbit(j)){
            res += c[cc][i][j];
        }
    }
    return res;
}
int main(){
    while(scanf("%d%d", &n, &m) != EOF)
    {
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= m; j++){
                scanf("%d", &maze[i][j]);
                update(i, j, maze[i][j], 1);
            }
        }
        scanf("%d", &q);
        while(q--){
            int op;
            scanf("%d", &op);
            if(op == 1){
                int x, y, c;
                scanf("%d%d%d", &x, &y, &c);
                update(x, y, maze[x][y], -1);
                maze[x][y] = c;
                update(x, y, maze[x][y], 1);
            }
            else{
                int x1, y1, x2, y2, c;
                scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &c);
                int ans = query(x2, y2, c) + query(x1 - 1, y1 - 1, c) - query(x2, y1 - 1, c) - query(x1 - 1, y2, c);
                printf("%d\n", ans);
            }
        }
    }
    return 0;
}