hdu1312

Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19354 Accepted Submission(s): 11772

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
思路：用到深搜，是其中一种比较简单的，分四个方向，标记一下，深搜出结果

#include<stdio.h>#include<string.h>#define N 22char a[N][N];int vis[N][N];int w,h;int move[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int ans;void dfs(int x,int y){    for(int i=0;i<4;i++)    {        int tx=x+move[i][0];        int ty=y+move[i][1];        if(tx>=0&&tx<h&&ty>=0&&ty<w&&a[tx][ty]=='.'&&!vis[tx][ty])        {            vis[tx][ty]=1;            ans++;            dfs(tx,ty);        }    }}int main(){    while(scanf("%d%d",&w,&h)&&w+h)    {         ans=0;         for(int i=0;i<h;i++)         {            scanf("%s",a[i]);        }        memset(vis,0,sizeof(vis));        for(int i=0;i<h;i++)        {            for(int j=0;j<w;j++)            {                if(a[i][j]=='@')                {                    vis[i][j]=1;                    ans++;                    dfs(i,j);                }            }           }              printf("%d\n",ans);    }    return 0;}