poj 1195 Mobile phones(二维树状数组)

题目链接：poj 1195 Mobile phones

题目大意：四种操作

• 0 s：清空(1,1)~(s,s)
• 1 x y a：在(x,y)点加上a
• 2 x1 y1 x2 y2：询问矩形(x1,y1)~(x2,y2)的和。
• 3：结束

解题思路：纯二维树状数组。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 1030;

#define lowbit(x) ((x)&(-x))
int fenw[maxn+5][maxn+5];

void add (int x, int y, int a) {
    for (int i = x; i <= maxn; i += lowbit(i)) {
        for (int j = y; j <= maxn; j += lowbit(j))
            fenw[i][j] += a;
    }
}

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

int main () {
    int type;
    int s, x, y, a, l, r;
    while (scanf("%d", &type) == 1 && type != 3) {
        if (type == 0) {
            scanf("%d", &s);
            for (int i = 1; i <= s; i++)
                for (int j = 1; j <= s; j++)
                    fenw[i][j] = 0;
        } else if (type == 1) {
            scanf("%d%d%d", &x, &y, &a);
            x++; y++;
            add(x, y, a);
        } else {
            scanf("%d%d%d%d", &x, &y, &l, &r);
            x++; y++; l++; r++;
            printf("%d\n", sum(l, r) - sum(x-1, r) - sum(l, y-1) + sum(x-1, y-1));
        }
    }
    return 0;
}