POJ 2155 二维树状数组（插块问点）

Matrix

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 24337 Accepted: 9013

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.

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).

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

二维树状数组的S[i][j]就是在从0 0 到i j这个子矩阵里的的所有元素的和，这里只有 0 1 所以%2，GG

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=1005;
int n;
int c

;
int lowbit(int x) //求最低位1的位置所表示的数
{
return x&(-x);
}
void update(int i,int j,int num)// 单点更新c[i][j] 加上num,
{
for(int x=i;x<=n;x+=lowbit(x))
for(int y=j;y<=n;y+=lowbit(y))
c[x][y]=(c[x][y]+num)%2;
}
int S(int i,int j)  //求c[i][j]的矩阵和
{
int sum=0;
for(int x=i;x>0;x-=lowbit(x))
for(int y=j;y>0;y-=lowbit(y))
sum=(sum+c[x][y])%2;
return sum;
}
int main(){
int T,t;
char s[2];
int x1,y1,x2,y2;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&t);
memset(c,0,sizeof(c));
while(t--){
scanf("%s",s);
if(s[0]=='C'){
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
update(x1,y1,1);//这里是重点，画图就可以明白了
update(x2+1,y2+1,1);//...
update(x1,y2+1,1);//....
update(x2+1,y1,1);//...
}
else {
scanf("%d%d",&x1,&y1);
printf("%d\n",S(x1,y1));
}
}
if(T)printf("\n");
}
return 0;
}