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POJ 1979 Red and Black

时间:2017-02-02 15:35:55      阅读:197      评论:0      收藏:0      [点我收藏+]
Red and Black
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 33405   Accepted: 18176

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) 
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

Japan 2004 Domestic
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 char map[25][25];
 6 bool vis[25][25]; 
 7 int n,m,sx,sy,sum;
 8 int dx[]={0,0,1,-1};
 9 int dy[]={1,-1,0,0};
10 void Dfs(int x,int y){
11     sum++;vis[x][y]=true;
12     for(int i=0;i<4;i++){
13         int nx=x+dx[i],ny=y+dy[i];
14         if(nx<=n&&nx>=1&&ny<=m&&ny>=1&&
15             map[nx][ny]==.&&!vis[nx][ny])
16         Dfs(nx,ny);
17     }
18 }
19 void init(){
20     memset(vis,false,sizeof(vis));
21     memset(map,0,sizeof(map));
22     sum=0;
23 }
24 int main()
25 {
26     while(1){
27         scanf("%d%d",&m,&n);
28         if(n==0&&m==0) break;
29         init();
30         for(int i=1;i<=n;i++)
31           for(int j=1;j<=m;j++){
32               cin>>map[i][j];
33               if(map[i][j]==@) sx=i,sy=j;
34           }
35         Dfs(sx,sy);
36         printf("%d\n",sum);
37     }
38     return 0;
39 }

 

POJ 1979 Red and Black

原文:http://www.cnblogs.com/suishiguang/p/6361201.html

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