POJ 1979 Red and Black （深搜）

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 23095 Accepted: 12467

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

Japan 2004 Domestic

简答的模板，

代码如下；

#include<stdio.h>#include<string.h>char a[22][22];int v[22][22];int sum;void dfs(int i,int j){if(a[i-1][j]=='.'&&v[i-1][j]==0){sum++;v[i-1][j]=1;dfs(i-1,j);}if(a[i][j-1]=='.'&&v[i][j-1]==0){sum++;v[i][j-1]=1;dfs(i,j-1);}if(a[i][j+1]=='.'&&v[i][j+1]==0){sum++;v[i][j+1]=1;dfs(i,j+1);}if(a[i+1][j]=='.'&&v[i+1][j]==0){sum++;v[i+1][j]=1;dfs(i+1,j);}}int main(){int n,m,i,j;while(~scanf("%d%d",&n,&m),n||m){getchar();sum=0;memset(v,0,sizeof(v));memset(a,'#',sizeof(a));for(i=0;i<m;i++){for(j=0;j<n;j++){scanf("%c",&a[i][j]);}getchar();}for(i=0;i<m;i++){for(j=0;j<n;j++){if(a[i][j]=='@'){v[i][j]=1;sum+=1;dfs(i,j);break;}}}printf("%d\n",sum);}return 0;}