hihocoder-Week 172--Matrix Sum

hihocoder-Week 172--Matrix Sum

 Matrix Sum

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

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 0

2. Sum x1 y1 x2 y1: Return the sum of every element Axy for x1 ≤ x ≤ x2, y1 ≤ y ≤ y2.

输入

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 

输出

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

样例输入

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 

样例输出

1
2
3 

题解：

　　二维树状数组

假设我们有二维数组A[1..N][1..N]，二维树状数组可以支持对A[][]进行如下操作： 
1. add(x, y, val): A[x][y] += val 
2. sum(x, y): 求和sum(A[1..x][1..y])

和一维情况类似，能支持以上两个操作实际就能支持任意修改A[x][y]的值以及求一个子矩阵A[a..b][c..d]的和。

二维树状数组以上两个操作的复杂度都是O(logNlogN)的。

二维树状数组BIT2[x][y]与A[][]的对应关系如下图：

直观理解就是x坐标和y坐标分别是一个一维树状数组，假设一维情况中BIT[x]对应的是A[i1], A[i2] ... A[ik]， BIT[y]对应的是A[j1], A[j2], ... A[jt]。那么BIT2[x][y] 相当于笛卡尔积 {i1, i2, ... ik} x {j1, j2, ... jt}：

BIT2][x][y] = ΣA[i][j] | {i in {i1 ... ik}且 j in {j1 ... jt}}

#include <cstdio>  
#include <cstdlib> 
#include <cstring> 

const int MAXN = 1000 + 10; 
const int MOD = 1e9 + 7; 

int n, m, mp[MAXN][MAXN]; 

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

void Add(int x, int y, int val){
	for(int i=x; i <= n; i += lowbit(i) ){
		for(int j=y; j <= n; j += lowbit(j) ){
			mp[i][j] += val; 
			mp[i][j] %= MOD; 
		}
	}
}

long long GetSum(int x1, int y1, int x2, int y2){
	long long ans1 = 0, ans2 = 0, ans3 = 0, ans4 = 0; 
	for(int i=x2; i>0; i-=lowbit(i)){
		for(int j=y2; j>0; j-=lowbit(j)){
			ans1 += mp[i][j]; 
			ans1 %= MOD;  
		}
	}
	for(int i=x1; i>0; i-=lowbit(i)){
		for(int j=y1; j>0; j-=lowbit(j)){
			ans2 += mp[i][j]; 
			ans2 %= MOD; 
		} 
	} 

	for(int i=x1; i>0; i-=lowbit(i)){
		for(int j=y2; j>0; j-=lowbit(j)){
			ans3 += mp[i][j]; 
			ans3 %= MOD; 
		}
	}
	for(int i=x2; i>0; i-=lowbit(i)){
		for(int j=y1; j>0; j-=lowbit(j)){
			ans4 += mp[i][j]; 
			ans4 %= MOD; 
		}
	} 
	return (ans1 + ans2 - ans3 - ans4 + MOD) % MOD; 
}

int main(){
	freopen("in.txt", "r", stdin); 

	int x1, x2, y1, y2, val; 
	long long ans; 
	char ch[10]; 

	while(scanf("%d %d", &n, &m) != EOF){  
		memset(mp, 0,  sizeof(mp)); 

		for(int i=0; i<m; ++i){
			getchar(); 
			scanf("%s", ch); 

			if(strcmp(ch, "Add") == 0){ 
				scanf("%d %d %d", &x1, &y1, &val);  
				Add(x1 + 1, y1 + 1, val);  

			}else if(strcmp(ch, "Sum") == 0){ 
				scanf("%d %d %d %d", &x1, &y1, &x2, &y2); 
				ans = GetSum(x1, y1, x2 + 1, y2 + 1); 

				printf("%lld
", ans ); 
			} 
		} 
	} 
	return 0; 
}