【BZOJ】1452: [JSOI2009]Count 树状数组

Description

Input

Output

Sample Input

Sample Output

1
2

HINT

  题解：

　　二维的树状数组啊+一维的颜色状态，然后直接做就好……实际上比照一维的树状数组就是多了一个for循环，然后查询操作的时候就相当于查询某一矩阵的大小，树状数组起到一个类似前缀和的作用。

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define lowbit(x) x&(-x)
using namespace std;
const int MAXN = 301;
int a[MAXN][MAXN], Tree[101][MAXN][MAXN];
int n, m;
void Modify(int x, int y, int s, int tar)
{
    int i, j;
    for (i = x; i <= n; i += lowbit(i))
        for (j = y; j <= m; j += lowbit(j))
            Tree[s][i][j] += tar;
}
inline int Query(int x, int y, int s)
{
    int i, j;
    int sum = 0;
    for (i = x; i; i -= lowbit(i))
        for (j = y; j; j -= lowbit(j))
            sum += Tree[s][i][j];
    return sum;
}
int main(int argc, char *argv[])
{
    int c, i, j, x, Q, op, y;
    int x1, y1, x2, y2;
    int ans;
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++)
        for (j = 1; j <= m; j++)
        {
        scanf("%d", &a[i][j]);
        Modify(i, j, a[i][j], 1);
        }
    scanf("%d", &Q);
    while (Q--)
    {
        scanf("%d", &op);
        if (op == 1)
        {
            scanf("%d%d%d", &x, &y, &c);
            Modify(x, y, a[x][y], -1);
            a[x][y] = c;
            Modify(x, y, a[x][y], 1);
        }
        else if (op == 2)
        {
            scanf("%d%d%d%d%d", &x1, &x2, &y1, &y2, &c);
            ans = 0;
            ans = Query(x1 - 1, y1 - 1, c) + Query(x2, y2, c) - Query(x1 - 1, y2, c) - Query(x2 , y1 - 1, c);
            printf("%d
", ans);
        }
    }
    return 0;
}