hdoj 1312 Red and Black

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 71   Accepted Submission(s) : 65

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：求@点最多能走多少步！#是墙，不能走！
搜索即可：
dfs：
#include<stdio.h>int cnt=0;int n,m;char a[22][22];void dfs(int x,int y){    if(a[x][y]=='#')    return ;    if(y>=1&&y<=n&&x<=m&&x>=1)    {        a[x][y]='#';        cnt++;     dfs(x+1,y);//四个方向搜索     dfs(x-1,y);    dfs(x,y+1);    dfs(x,y-1);   }}int main(){    int i,j;    while(scanf("%d%d",&n,&m),n|m)    {        cnt=0;        int x,y;            for(i=1;i<=m;i++)            {            getchar();              for(j=1;j<=n;j++)                {                    scanf("%c",&a[i][j]);                    if(a[i][j]=='@')//确定起点                     {                        x=i;                        y=j;                    }                }            }            dfs(x,y);      printf("%d\n",cnt);    }}