TLE 的同学们注意了。下标从0开始。附上测试数据。

in
0 1
1 0 0 28766
2 0 0 0 0
1 0 0 -2826
2 0 0 0 0
1 0 0 -587
2 0 0 0 0
1 0 0 1017
2 0 0 0 0
1 0 0 -17234
2 0 0 0 0
1 0 0 19618
2 0 0 0 0
1 0 0 -2237
2 0 0 0 0
1 0 0 -9192
2 0 0 0 0
1 0 0 -11132
2 0 0 0 0
1 0 0 1842
2 0 0 0 0
3

out
28766
25940
25353
26370
9136
28754
26517
17325
6193
8035

过了这组应该就没有问题了。

//////////////My AC Code///////////////////
/*  641MS	C++  */
/*  1079MS	G++  */
#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const int MAXN = 1050;
int s[MAXN][MAXN];
int lowbit[MAXN];
int n;

int sum(int x, int y)
{
    int res = 0;
    for(int i = x; i > 0; i -= lowbit[i])
        for(int j = y; j > 0; j -= lowbit[j])
            res += s[i][j];
    return res;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int op;
    int x, y, v;
    int x1, y1, x2, y2;
    for(int i = 1; i <= MAXN; i++)
    {
        lowbit[i] = i&-i;
    }
    while(1)
    {
        scanf("%d", &op);
        if(op == 1)
        {
            scanf("%d %d %d", &x, &y, &v);
            x++; y++;
            for(int i = x; i <= n; i += lowbit[i])
                for(int j = y; j <= n; j += lowbit[j])
                    s[i][j] += v;
        }
        else if(op == 2)
        {
            scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
            x1++; y1++; x2++; y2++;
            int ans = sum(x2, y2)+sum(x1-1, y1-1)-sum(x1-1, y2)-sum(x2, y1-1);
            printf("%d\n", ans);
        }
        else if(op == 0)
        {
            scanf("%d", &n);
            memset(s, 0, sizeof(s));
        }
        else
            break;
    }
    return 0;
}

• Re:二维树状数组 TLE 的同学们注意了 ！ 下标从0开始。 附上测试数据。 (2083) Gazer_Fri 2016-07-26 15:10:29
• Re:二维树状数组 TLE 的同学们注意了 ！ 下标从0开始。 附上测试数据。 (2074) liang23333 2016-09-01 19:43:04