POJ2155 Matrix

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25080		Accepted: 9293

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

 

二维树状数组果然比二维线段树简单多了

讲一下二维树状数组

其实我也不清楚多出来的一维怎么做

但既然多套了一重循环就算作是二维了

文字说不清，自己仿照一维画一个图就明白了

再说这道题

假设只有一维，我们可以用树状数组维护一个差分数组，区间首尾打标记+1，求和即可

那么推广到二维，把维护差分数组的方式看成打一个标记

四个点+1，对询问求一遍和模2

尹神的办法zrl说可以推广，而这种办法只对01有效

就是说对于正常的差分数组，区间修改应该是首加尾减

到了二维应该这样维护

-1 +1

+1 -1

就这样吧

 

——by隔壁老司机RLQ

 

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=1200;
int t[mxn][mxn];
int n,m;
int lowbit(int x){return x&-x;}
int an;
void add(int x,int y,int v){
    while(x<=n){
        int tmp=y;
        while(tmp<=n){ t[x][tmp]+=v; tmp+=lowbit(tmp);}
        x+=lowbit(x);
    }
}
int sum(int x,int y){
    int res=0;
    while(x){
        int tmp=y;
        while(tmp){    res+=t[x][tmp];     tmp-=lowbit(tmp);}
        x-=lowbit(x);
    }
    return res;
}
int ask(int x1,int y1,int x2,int y2){
    return sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1,y2-1);
}
int main(){
    int T;
    scanf("%d",&T);
    int i,j;
    char ch[5];
    int x1,y1,x2,y2;
    while(T--){
        memset(t,0,sizeof t);
        scanf("%d%d",&n,&m);
        an=n+5;
        for(i=1;i<=m;i++){
            scanf("%s",ch);
            if(ch[0]=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,1);
                add(x1,y2+1,-1);
                add(x2+1,y1,-1);
                add(x2+1,y2+1,1);
            }
            else{
                scanf("%d%d",&x1,&y1);
                printf("%d\n",sum(x1,y1)%2);
            }
        }
        printf("\n");
    }
    return 0;
}