hdu 1312 Red and Black

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Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10133 Accepted Submission(s): 6321

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

....#.

.....#

......

#@...#

.#..#.

0 0

Sample Output

Source

Asia 2004, Ehime (Japan), Japan Domestic

PS:简单题

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<cstring> 6 #include<queue> 7 using namespace std; 8 int n,m; 9 int xi,yj;10 char map[25][25];11 int vis[25][25];12 int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//上下左右13 struct node{14     int x;15     int y;16 };17 18 bool check(int x,int y)19 {20     if(x>=1&&x<=n && y>=1&&y<=m && map[x][y]=='.' && !vis[x][y])21     return true;22     return false;23 }24 25 int bfs()26 {27     memset(vis,0,sizeof(vis));28     int i,sum=1;29     node p,q;30     queue<struct node>Q;31     p.x=xi;32     p.y=yj;33     Q.push(p);34     while(!Q.empty())35     {36         p=Q.front();37         Q.pop();38 39         for(i=0;i<4;i++)40         {41             q.x=p.x+dir[i][0];42             q.y=p.y+dir[i][1];43             if(check(q.x,q.y))44             {45                 vis[q.x][q.y]=1;46                 sum++;47                 Q.push(q);48             }49         }50     }51     return sum;52 }53 54 55 int main()56 {57     int i,j;58     int count;59     //freopen("in.txt","r",stdin);60     while(~scanf("%d%d",&m,&n))61     {62 63         if(!n&&!m)64         break;65         for(i=1;i<=n;i++)66         {67             getchar();68             for(j=1;j<=m;j++)69             {70                 scanf("%c",&map[i][j]);71                 if(map[i][j]=='@')72                 {73                     xi=i;74                     yj=j;75                 }76             }77         }78         count=bfs();79         printf("%d\n",count);80     }81     return 0;82 }

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