poj 2155 Matrix

poj 2155 Matrix

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 24715 Accepted: 9156
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

理解这个需要炒鸡熟悉树状数组熟悉到你的输入法都知道szsz是树状数组的地步

步骤：

http://blog.csdn.net/ljd4305/article/details/10101535

理解“其实Update()和Getsum()这两个函数是相同的”

http://blog.csdn.net/lingyunjinzhu/article/details/8110444

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int lowbit(int x){return x&(-x);}int cc[1010][1010];int n;void change(int a,int b,int c,int d){for(int i = c;i > 0;i -= lowbit(i)){for(int j = d;j > 0;j -= lowbit(j)){cc[i][j] += 1;}}for(int i = (a-1);i > 0;i -= lowbit(i)){for(int j = (b-1);j > 0;j -= lowbit(j)){cc[i][j] += 1;}}for(int i = (a-1);i > 0;i -= lowbit(i)){for(int j = d;j > 0;j -= lowbit(j)){cc[i][j] -= 1;}}for(int i = c;i > 0;i -= lowbit(i)){for(int j = (b-1);j > 0;j -= lowbit(j)){cc[i][j] -= 1;}}}int _01(int x,int y){int ans = 0;for(int i = x;i <= n;i += lowbit(i)){for(int j = y;j <= n;j += lowbit(j)){ans += cc[i][j];}}return (ans%2);}int main(){int T;int F = 0;scanf("%d",&T);while(T--){if(F) printf("\n");F++;int qn;scanf("%d%d",&n,&qn);memset(cc,0,sizeof(cc));while(qn--){char ch[10];int a,b,c,d;scanf("%s",ch);if(ch[0] == 'Q'){scanf("%d%d",&a,&b);printf("%d\n",_01(a,b));}else{scanf("%d%d%d%d",&a,&b,&c,&d);change(a,b,c,d);}}}}