POJ

Matrix

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

题意：一个n*n区域，m次操作 C 代表我们将区域转换状态，Q 为查询当前这个点状态
思路：我们累加转换次数 偶数就还是为0 奇数为1 关键在于我们如何转换x,y,x1,y1这个区域。
我们在翻转x,y,x1,y1的过程中我们也将x1,y1,n,n翻转了我们只需要将其他区域翻转回来即可 详细见代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define pb push_back
#define pll pair<int,int>
#define mp make_pair
#define ll long long
const int maxn = 1000+10;
int tree[maxn][maxn];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int y)
{
    for(int i=x; i<=n; i+=lowbit(i))
        for(int j=y; j<=n; j+=lowbit(j))
            tree[i][j]++;
}
int query(int x,int y)
{
    int ans=0;
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            ans+=tree[i][j];
    return ans;
}
int main()
{
    int t,m,x1,x2,y1,y2;
    scanf("%d",&t);
    while(t--)
    {
        char ch[10];
        memset(tree,0,sizeof(tree));
        scanf("%d %d",&n,&m);
        while(m--)
        {
            scanf("%s",ch);
            if(ch[0]=='C')
            {
                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                add(x1,y1);
                add(x1,y2+1);
                add(x2+1,y1);
                add(x2+1,y2+1);
            }
            else
            {
                scanf("%d %d",&x1,&y1);
                printf("%d
",query(x1,y1)&1);
            }
        }
        printf("
");
    }

}

 PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~