Hdu4146

Flip Game

Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1842    Accepted Submission(s): 607

Problem Description

Flip game is played on a square N*N field with two-sided pieces placed on each of its N^2 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. The rows are numbered with integers from 1 to N upside down; the columns are numbered with integers from 1 to N from the left to the right. Sequences of commands (x_i, y_i) are given from input, which means that both pieces in row x_i and pieces in column y_i will be flipped (Note that piece (x_i, y_i) will be flipped twice here). Can you tell me how many white pieces after sequences of commands?
Consider the following 4*4 field as an example:

bwww
wbww
wwbw
wwwb

Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up.
Two commands are given in order: (1, 1), (4, 4). Then we can get the final 4*4 field as follows:

bbbw
bbwb
bwbb
wbbb

So the answer is 4 as there are 4 white pieces in the final field.

 

Input

The first line contains a positive integer T, indicating the number of test cases (1 <= T <= 20).
For each case, the first line contains a positive integer N, indicating the size of field; The following N lines contain N characters each which represent the initial field. The following line contain an integer Q, indicating the number of commands; each of the following Q lines contains two integer (x_i, y_i), represent a command (1 <= N <= 1000, 0 <= Q <= 100000, 1 <= x_i, y_i <= N).

 

Output

For each case, please print the case number (beginning with 1) and the number of white pieces after sequences of commands.

 

Sample Input

24bwwwwbwwwwbwwwwb21 14 44wwwwwwwwwwwwwwww11 1

 

Sample Output

Case #1: 4Case #2: 10

题意：给出一个有黑子和白子的图，有多次转化，输入X,Y就是把x行和y列的翻转。

求最后白子的数量。

1. #include<stdio.h>  
2. #include<string.h>  
3. int main()  
4. {  
5.     int s,n,m,x,y,i,j,count,sum=1;  
6.     char str[1001][1001];  
7.     int a[1001],b[1001];  
8.     scanf("%d",&s);  
9.     while(s--)  
10.     {  
11.         memset(a,0,sizeof(a));  
12.         memset(b,0,sizeof(b));  
13.         count=0;  
14.         scanf("%d",&n);  
15.         for(i=0;i<n;i++)  
16.         {  
17.             scanf("%s",str[i]);//**这里需要解释下，二维数组输入str[i][j]时，首先读入的是str[0][j]**//  
18.         }  
19.         scanf("%d",&m);  
20.         for(i=0;i<m;i++)  
21.         {  
22.             scanf("%d %d",&x,&y);//**代表第x行，第y列**//  
23.             x--,y--;//**因为数组下表是从0开始的//**printf("%d %d\n",x,y);**//*//  
24.             a[x]=1-a[x];//**x的横坐标改变**//  
25.             b[y]=1-b[y];//**y的纵坐标改变**//  
26.         }  
27.         for(i=0;i<n;i++)  
28.         {  
29.             for(j=0;j<n;j++)  
30.             {  
31.                 if(a[i]+b[j]==1)//**判断改变次数的奇偶性<span style="font-size: 16px;">**//</span>  
32.                 {  
33.                     if(str[i][j]=='b') count++;  
34.                 }  
35.                 else  
36.                 {  
37.                     if(str[i][j]=='w') count++;  
38.                 }  
39.             }  
40.         }  
41.         printf("Case #%d: %d\n",sum++,count);  
42.     }  
43.     return 0;  
44. }