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Red and Black(dfs水)

时间:2016-02-22 15:50:23      阅读:152      评论:0      收藏:0      [点我收藏+]

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15186    Accepted Submission(s): 9401


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
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
 


Sample Output
45 59 6 13
 
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<string>
 5 using namespace std;
 6 #define N 25
 7 char mp[N][N];
 8 int go[4][2] = {{1,0},{-1,0},{0,-1},{0,1}};
 9 
10 int n,m;
11 bool ck(int x, int y){
12     if(x<n&&x>=0&&y<m&&y>=0) return true;
13     else return false;
14 }
15 bool vis[N][N];
16 void dfs(int x, int y, int &sum)
17 {
18     vis[x][y] = 1;
19     bool fl = 0;
20     for(int i = 0; i < 4; i++){
21         int xx = x + go[i][0];
22         int yy = y + go[i][1];
23         if(ck(xx,yy)&&!vis[xx][yy]&&mp[xx][yy]==.){
24             fl = 1;
25             sum = sum+1;
26             dfs(xx,yy,sum);
27         }
28     }
29     if(fl==0) return;
30 }
31 int main()
32 {
33     while(~scanf("%d%d",&m,&n))
34     {
35         int x, y;
36         if(n==0&&m==0) break;
37         getchar();
38         for(int i = 0; i < n; i++){
39             for(int j = 0; j < m; j++){
40                 scanf("%c",&mp[i][j]);
41                 if(mp[i][j]==@){ x = i; y = j; }
42             }
43             getchar();
44         }
45         memset(vis,0,sizeof(vis));
46         int sum = 0;
47         dfs(x,y,sum);
48         printf("%d\n",sum+1);
49     }
50     return 0;
51 }

 

Red and Black(dfs水)

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原文地址:http://www.cnblogs.com/shanyr/p/5207020.html

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