(HihoCoder - 1336)Matrix Sum

最新推荐文章于 2020-02-19 13:46:30 发布

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本文介绍了一种使用二维树状数组解决矩阵操作问题的方法。针对给定的n*n矩阵，通过addxyv指令在指定位置增加数值，并通过sumx1y1x2y2指令求解指定矩形区域内的元素总和。采用二维树状数组进行高效更新和查询，以应对大量操作。

(HihoCoder - 1336)Matrix Sum

You are given an N × N matrix. At the beginning every element is 0. Write a program supporting 2 operations:

Add x y value: Add value to the element Axy. (Subscripts starts from 0
1. Sum x1 y1 x2 y1: Return the sum of every element Axy for x1 ≤ x ≤ x2, y1 ≤ y ≤ y2.

Input

The first line contains 2 integers N and M, the size of the matrix and the number of operations.

Each of the following M line contains an operation.

1 ≤ N ≤ 1000, 1 ≤ M ≤ 100000

For each Add operation: 0 ≤ x < N, 0 ≤ y < N, -1000000 ≤ value ≤ 1000000

For each Sum operation: 0 ≤ x1 ≤ x2 < N, 0 ≤ y1 ≤ y2 < N

Output

For each Sum operation output a non-negative number denoting the sum modulo 109+7.

Sample Input

5 8
Add 0 0 1
Sum 0 0 1 1
Add 1 1 1
Sum 0 0 1 1
Add 2 2 1
Add 3 3 1
Add 4 4 -1
Sum 0 0 4 4

Sample Output

1
2
3

题目大意：对于一个n*n的二维数组，对其进行m次操作，add x y v 指在(x,y)处加v，sum x1 y1 x2 y2 询问在左下角为(x1,y1)右上角为(x2,y2)的矩形里面元素的和。

思路：二维树状数组模板里，用二维数组数组来维护这一系列操作。

#include<cstdio>
#include<cstring>
using namespace std;

typedef long long LL;
const LL mod=1e9+7;
const int maxn=1005;
LL c[maxn][maxn];
char op[5];

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

void update(int x,int y,int d)
{
    for(int i=x;i<maxn;i+=lowbit(i))
        for(int j=y;j<maxn;j+=lowbit(j)) c[i][j]=(c[i][j]+d)%mod;
}

LL getsum(int x,int y)
{
    LL sum=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j)) sum=(sum+c[i][j])%mod;
    return sum;
}

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(c,0,sizeof(c));
        while(m--)
        {
            scanf("%s",op);
            if(op[0]=='A')
            {
                int x,y,v;
                scanf("%d%d%d",&x,&y,&v);
                x++,y++;
                update(x,y,v);//+1的目的是为了排除0而避免死循环 
            }
            else
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                x1++,y1++,x2++,y2++;
                LL ans=((getsum(x2,y2)-getsum(x1-1,y2)-getsum(x2,y1-1)+getsum(x1-1,y1-1))%mod+mod)%mod;
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}