poj2155 Matrix

一道二维树状数组题

这题要用差分的思想

将x1y1x2y2区间取反，就将x1,y1；x2+1,y1；x1，y2+1；x2+1,y2+1；加1

查询就直接输出，注意mod2

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 34294		Accepted: 12409

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

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int maxn = 1e3+10;

int t,n,k,tree[maxn][maxn];

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

void add(int x,int y,int delta){
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j)){
            tree[i][j]+=delta;
        }
}

int getsum(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;
}

char getc(){
    char ch=getchar();
    while(ch!='Q'&&ch!='C') ch=getchar();
    return ch;
}

int main(){
    scanf("%d",&t);
    while(t--){
        memset(tree,0,sizeof(tree));
        scanf("%d%d",&n,&k);
        for(int i=1;i<=k;i++){
            char ch=getc();
            if(ch=='C'){
                int x1,y1,x2,y2;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{
                int x1,y1;
                scanf("%d%d",&x1,&y1);
                printf("%d
",getsum(x1,y1)%2);
            }
        }
        if(t) printf("
");//题目要求，要加
    }
    return 0;
}