POJ2155:Matrix

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 32921		Accepted: 11933

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

POJ Monthly,Lou Tiancheng

 

题目大意：给你一个初始都为0的n*n的矩阵，Cx1,y1,x2,y2,对这个小矩阵取反，即0->1,1->0,Qx1,y1,查询这个点为1还是0？

二维树状数组裸题

支持区间修改和单点查询

 

由于不能维护每个点的状态，但可以维护每个点的修改次数，若修改次数%2==0，则相当于不发生改变。

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<vector>
using namespace std;

void in(int &x){
    register char c=getchar();x=0;int f=1;
    while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    x*=f;
}

const int N=1005;
int bit[N][N],t,m,n;

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

void add(int x,int y,int z){
    int ty=y;
    while(x<=n){
        y=ty;
        while(y<=n)
            bit[x][y]+=z,y+=lowbit(y);
        x+=lowbit(x);
    }
}

void change_interval(int x1,int y1,int x2,int y2){
    add(x1,y1,1);
    add(x1,y2+1,-1);
    add(x2+1,y1,-1);
    add(x2+1,y2+1,1);
}

int ask(int x,int y){
    int sum=0,ty=y;
    while(x>0){
        y=ty;
        while(y>0)
            sum+=bit[x][y],y-=lowbit(y);
        x-=lowbit(x);
    }return sum;
}
int main()
{
    in(t);
    while(t--){
        memset(bit,0,sizeof(bit));
        in(n);in(m);
        for(int i=1,x1,y1,x2,y2;i<=m;i++){
            char p;
            scanf("%c",&p);in(x1);in(y1);
            if(p=='C'){
                in(x2);in(y2);
                change_interval(x1,y1,x2,y2);
            }else{
                printf("%d
",ask(x1,y1)%2);
            }
        }printf("
");//注意，否则会PE
    }return 0;
}