hihoCoder 1336：Matrix Sum（二维树状数组）

Matrix Sum

Time limit:1000 ms Memory limit:256MB

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Problem Description

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

1. Add x y value: Add value to the element Axy. (Subscripts starts from 02. Sum x1 y1 x2 y2: 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 8Add 0 0 1Sum 0 0 1 1Add 1 1 1Sum 0 0 1 1Add 2 2 1Add 3 3 1Add 4 4 -1Sum 0 0 4 4 

Sample Output

123 

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题意：

给你一个N*N二维矩阵，初始值时都为0，有两种操作

1：单点修改值
2：给出一个子矩阵的左上和右下角的坐标，询问一个子矩形范围内的值的和.

解题思路：

还是看官方的题解吧。。。《Matrix Sum》题目分析

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Code：

#include <iostream>#include <cstdio>#include <cstring>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long LL;const int maxn=1000+5,          mod=1e9+7;LL BIT[maxn][maxn];int N,m;int lowbit(int x){    return x&(-x);}void add(int x,int y,int val){    for(int i=x;i<maxn;i+=lowbit(i))    {        for(int j=y;j<maxn;j+=lowbit(j))        {            BIT[i][j]=(BIT[i][j]+val)%mod;        }    }}int sum(int x,int y){    LL ret=0;    for(int i=x;i>0;i-=lowbit(i))    {        for(int j=y;j>0;j-=lowbit(j))        {            ret=(ret+BIT[i][j])%mod;        }    }    return (int)ret;}int main(){    mem(BIT,0);    scanf("%d%d",&N,&m);    char op[10];    for(int i=0;i<m;i++)    {        scanf("%s",op);        if(op[0]=='A')        {            int x,y,val;            scanf("%d%d%d",&x,&y,&val);            x++,y++;            add(x,y,val);        }        else        {            int x1,y1,x2,y2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            x1++,x2++,y1++,y2++;            int res=(sum(x2,y2)-sum(x1-1,y2)-sum(x2,y1-1)+sum(x1-1,y1-1)+mod)%mod;            printf("%d\n",res);        }    }    return 0;}