poj 1979 Red and Black (简单裸搜索)

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 28661 Accepted: 15616

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

4559613

Source

题目链接：http://poj.org/problem?id=1979

题目大意：'.'表示黑点，'#'表示红点，'@'表示一个人的起始位置，人只能在黑点上移动且可以往上下左右四个方向移动，求这个人最多能走的格子数。

解题思路：裸的深搜。

代码如下：

#include <cstdio>#include <cstring>int n,m;char a[25][25];int vis[25][25];int ans;int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};void dfs(int x,int y){for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx>=n || ny>=m || nx<0 ||ny<0 || vis[nx][ny])continue;if(a[nx][ny]=='#')continue;vis[nx][ny]=1;dfs(nx,ny);ans++;}}int main(void){while(scanf("%d%d",&m,&n)!=EOF && (n+m)){int sx,sy;ans=0;memset(vis,0,sizeof(vis));for(int i=0;i<n;i++){scanf("%s",a[i]);for(int j=0;j<m;j++)if(a[i][j]=='@'){sx=i;sy=j;}}vis[sx][sy]=1;a[sx][sy]='.';dfs(sx,sy);printf("%d\n",ans+1 );}}