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Matrix(poj2155)

2024-08-14 05:32:05   222人阅读

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 25139 Accepted: 9314

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

Source

思路：二维树状数组；

http://download.csdn.net/detail/lenleaves/4548401

这个解释的很好；

 1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<stdlib.h> 5 #include<queue> 6 #include<string.h> 7 using namespace std; 8 int bit[1005][1005]; 9 int lowbit(int x)10 {11         return x&(-x);12 }13 void add(int x,int y)14 {15         int i,j;16         for(i = x; i <= 1000; i+=lowbit(i))17         {18                 for(j = y; j <= 1000; j+=lowbit(j))19                 {20                         bit[i][j]+=1;21                         bit[i][j]%=2;22                 }23         }24 }25 int ask(int x,int y)26 {27         int i,j;28         int sum = 0;29         for(i = x; i > 0; i-=lowbit(i))30         {31                 for(j = y; j > 0; j-=lowbit(j))32                 {33                         sum += bit[i][j];34                 }35         }36         return sum%2;37 }38 int main(void)39 {40         int T;41         scanf("%d ",&T);42         while(T--)43         {44                 memset(bit,0,sizeof(bit));45                 int i,j;46                 int N,q;47                 scanf("%d %d ",&N,&q);48                 char a[10];49                 while(q--)50                 {51                         scanf("%s",a);52                         int x,y,x1,y1;53                         if(a[0] == ‘C‘)54                         {55                                 scanf("%d %d %d %d",&x,&y,&x1,&y1);56                                 add(x,y);57                                 add(x1+1,y1+1);58                                 add(x,y1+1);59                                 add(x1+1,y);60                         }61                         else62                         {63                                 scanf("%d %d",&x,&y);64                                 int ac = ask(x,y);65                                 printf("%d\n",ac);66                         }67                 }68                 printf("\n");69         }70         return 0;71 }

 

Matrix(poj2155)


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