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poj 1979 Red and Black(dfs水题)

时间:2015-08-28 21:19:50      阅读:143      评论:0      收藏:0      [点我收藏+]

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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 cant 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) 
The end of the input is indicated by a line consisting of two zeros. 

 

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

 

Source

 
 
 
技术分享
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 #define N 26
 6 int n,m;
 7 char mp[N][N];
 8 int x,y;
 9 int vis[N][N];
10 int ans;
11 int dirx[]={0,0,-1,1};
12 int diry[]={-1,1,0,0};
13 void dfs(int sx,int sy){
14     for(int i=0;i<4;i++){
15         int xx=sx+dirx[i];
16         int yy=sy+diry[i];
17         if(vis[xx][yy]) continue;
18         if(mp[xx][yy]==#) continue;
19         if(xx<0 || xx>=n || yy<0 || yy>=m) continue;
20         vis[xx][yy]=1;
21         ans++;
22         dfs(xx,yy);
23     }
24 }
25 int main()
26 {
27     while(scanf("%d%d",&m,&n)==2 && n+m){
28         for(int i=0;i<n;i++){
29             scanf("%s",mp[i]);
30             for(int j=0;j<m;j++){
31                 if(mp[i][j]==@){
32                     x=i;y=j;
33                 }
34             }
35         }
36         memset(vis,0,sizeof(vis));
37         ans=1;
38         vis[x][y]=1;
39         dfs(x,y);
40         printf("%d\n",ans);
41 
42     }
43     return 0;
44 }
View Code

 

poj 1979 Red and Black(dfs水题)

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原文地址:http://www.cnblogs.com/UniqueColor/p/4767670.html

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