hdoj1312Red and Black（递归）

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13136    Accepted Submission(s): 8136

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

4559613

 题意：某人在@处为起点（也包括@点）#为墙，点（.）为通路，问最多能走多远统计能走几个点（加上@这个点）

#include<stdio.h>char a[50][50];int n,m,count;void fun(int x,int y){if(a[x][y]=='#')//看看是不是墙 return ;if(x>m||y>n||x<1||y<1)//在范围内 return ;count++;a[x][y]='#';//已经查到得这个点就要变成墙，否则的话 就变成无限的查找了 fun(x-1,y);//查上下左右四个点 fun(x+1,y);fun(x,y+1);fun(x,y-1);}int main(){int i,j,b,c;while(~scanf("%d %d",&n,&m))//n行m列 {count=0;if(n==0||m==0)return 0;for(i=1;i<=m;i++){getchar();for(j=1;j<=n;j++){scanf("%c",&a[i][j]);//找出@点在几行几列 if(a[i][j]=='@')b=i,c=j;}}fun(b,c);//查一下在这个点周围还有几个点 printf("%d\n",count);}return 0;}