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Red and Black(水)

时间:2015-07-03 15:45:42      阅读:129      评论:0      收藏:0      [点我收藏+]

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12138    Accepted Submission(s): 7554

Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can‘t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.
 
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.‘ - a black tile
‘#‘ - a red tile
‘@‘ - a man on a black tile(appears exactly once in a data set)
 
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
 
Sample Input
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Sample Output
45
59
6
13
 1 #include <iostream>
 2 #include <cstdio>
 3 using namespace std;
 4 char a[21][21];
 5 int m,n;
 6 int count=0;
 7 void dfs(int x,int y)
 8 {
 9     if(a[x][y]==.&&x<n&&x>=0&&y<m&&y>=0)
10     {
11         count++;
12         a[x][y]=#;
13         dfs(x+1,y);
14         dfs(x,y+1);
15         dfs(x,y-1);
16         dfs(x-1,y);
17     }
18 }
19 int main()
20 {
21     freopen("in.txt","r",stdin);
22     while(scanf("%d%d",&m,&n)&&m!=0&&n!=0)
23     {
24         count=0;
25         int i,j,x,y;
26         for(i=0;i<n;i++)
27         {
28             for(j=0;j<m;j++)
29             {
30                 cin>>a[i][j];
31                 if(a[i][j]==@)
32                 {
33                     x=i;y=j;
34                 }
35             }
36         }
37         a[x][y]=.;
38         dfs(x,y);
39         cout<<count<<endl;
40     }
41 }

 

 
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Red and Black(水)

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原文地址:http://www.cnblogs.com/a1225234/p/4618518.html

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