（HDU

（HDU - 1312）Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 21462 Accepted Submission(s): 13072

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

题目大意：一个人从@处出发,’.’可以走，‘#’不能走，问最多可以走过多少个‘.’。

思路：典型的迷宫题，只不过这里要求的是最多的可以走的步数，由于数据较小，dfs便可解决，不用回溯。

#include<cstdio>using namespace std;const int maxn=25;char a[maxn][maxn];const int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int w,h,cnt;int dfs(int x,int y){    cnt++;    a[x][y]='#';    for(int i=0;i<4;i++)    {        int nx=x+dir[i][0];        int ny=y+dir[i][1];        if(a[nx][ny]=='.'&&nx>=0&&nx<h&&ny>=0&&ny<w)            dfs(nx,ny);    }    return cnt;}int main(){    while(scanf("%d%d",&w,&h)!=EOF&&(w||h))    {        for(int i=0;i<h;i++) scanf("%s",a[i]);        int sx,sy;        bool flag=false;        for(int i=0;i<h;i++)        {            for(int j=0;j<w;j++)                if(a[i][j]=='@')                {                    sx=i;                    sy=j;                    flag=true;                    break;                }               if(flag) break;         }        cnt=0;        printf("%d\n",dfs(sx,sy));      }    return 0;}