poj 1979 Red and Black -- dfs

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 29614 Accepted: 16095

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

题意是从@点出发，然后遍历整个图，‘.’表示可以到达的地方，求最多可以走多少个‘.’，初始的点也算一个，注意sum=1初始化

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;char map[25][25];int vis[25][25];int n,m,num;int dr[4]={0,1,-1,0};int dc[4]={-1,0,0,1};void dfs(int x,int y){int i;for(i=0;i<4;i++){int dx=x+dr[i];int dy=y+dc[i];if(dx>=0&&dx<n&&dy>=0&&dy<m&&!vis[dx][dy]&&map[dx][dy]=='.'){vis[dx][dy]=1;num++;dfs(dx,dy);}}}int main(){int i,j,x,y;//freopen("test.txt","r",stdin);while(~scanf("%d %d",&m,&n)){if(m==0&&n==0)break;char c[25];for(i=0;i<n;i++){scanf("%s",&c);for(j=0;j<m;j++){map[i][j]=c[j];if(map[i][j]=='@'){x=i;y=j;}}}//printf("%d %d\n",x,y);memset(vis,0,sizeof(vis));num=1;dfs(x,y);printf("%d\n",num);}return 0;}