POJ-1195-Mobile phones

链接：https://vjudge.net/problem/POJ-1195

题意：

给一个S*S的矩阵，有两种操作，给(x,y)位置增加一个值，和求一个内部矩形的和。

思路：

二维树状数组，先对每行来一个一维的树状数组，

再对行来一个树状数组

代码：

#include <iostream>
#include <memory.h>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <queue>
#include <string>
#include <stack>
#include <iterator>
#include <stdlib.h>
#include <time.h>
#include <assert.h>

using namespace std;
typedef long long LL;

const int MAXN = 2000;
int c[MAXN][MAXN];
int n, op;

int Lowbit(int x)
{
    return x&(-x);
}

void Update(int x, int y, int v)
{
    for (int i = x;i <= n;i += Lowbit(i))
    {
        for (int j = y;j <= n;j += Lowbit(j))
        {
            c[i][j] += v;
        }
    }
}

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 += c[i][j];
        }
    }
    return res;
}

int main()
{
    scanf("%d%d", &op, &n);
    while (~scanf("%d", &op))
    {
        if (op == 3)
            break;
        if (op == 1)
        {
            int x, y, v;
            scanf("%d%d%d", &x, &y, &v);
            x++, y++;
            Update(x, y, v);
        }
        if (op == 2)
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            x1++, y1++, x2++, y2++;
            printf("%d\n", Sum(x2, y2) - Sum(x1-1, y2) - Sum(x2, y1-1) + Sum(x1-1, y1-1));
        }
    }

    return 0;
}