Matrix（二维树状数组）入门第一题

题目链接：http://poj.org/problem?id=2155

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 34697		Accepted: 12542

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

Source

题目大意：T组样例 ，N*N的矩阵 M次操作  为C时  修改矩阵(x1,y1) (x2,y2)的元素 1变0  0变1   为Q时  询问坐标为（x1,y1）的位置 是什么

思路：这是第一道树状数组二维的题，说实话并没有想到怎么写，看了人家代码，果然牛逼。 复杂度应该是M*(log N)*(log N)吧 ，其实有两种写法，一种是往后存储 一种是往前存储

看代码：

1.往前存储的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const LL INF=1e9+7;
const int maxn=1e3+50;
int c[maxn][maxn];
int N,M;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y)
{
    for(int i=x;i>0;i-=lowbit(i))
    {
        for(int j=y;j>0;j-=lowbit(j)) c[i][j]++;
    }
}
int sum(int x,int y)
{
    int ret=0;
    for(int i=x;i<=N;i+=lowbit(i))
    {
        for(int j=y;j<=N;j+=lowbit(j)) ret+=c[i][j];
    }
    return ret%2;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));

        char c;
        scanf("%d%d",&N,&M);
        while(M--)
        {
            getchar();
            scanf("%c",&c);
            if(c=='C')
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                
                //这是精髓了  其实包围起来的就是我们所要修改的矩阵  如果想不通的话 看一下一维的是怎么存储区间问题的
                update(x1-1,y1-1);
                update(x1-1,y2);
                update(x2,y1-1);
                update(x2,y2);
            }
            else
            {
                int x1,y1;
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1));

            }

        }
        if(T!=0) printf("\n");
    }
    return 0;
}

2.往后存储的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL mod=1e9+7;
const LL INF=1e9+7;
const int maxn=1e3+50;
int c[maxn][maxn];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y)
{
    for(int i=x;i<1005;i+=lowbit(i))
    {
        for(int j=y;j<1005;j+=lowbit(j)) c[i][j]++;
    }
}
int sum(int x,int y)
{
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i))
    {
        for(int j=y;j>0;j-=lowbit(j)) ret+=c[i][j];
    }
    return ret%2;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(c,0,sizeof(c));
        int N,M;
        char c;
        scanf("%d%d",&N,&M);
        while(M--)
        {
            getchar();
            scanf("%c",&c);
            if(c=='C')
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                update(x1,y1);
                update(x1,y2+1);
                update(x2+1,y1);
                update(x2+1,y2+1);
            }
            else
            {
                int x1,y1;
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1));

            }

        }
        if(T!=0) printf("\n");
    }
    return 0;
}