poj1979 Red and Black（DFS）

Red and Black Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 25797   Accepted: 13967 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 45 59 6 13 Source Japan 2004 Domestic

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直接附上代码吧。很简单的深搜

#include <stdio.h>
#include <string.h>
char map[25][25];
int dir[4][2]={1,0,-1,0,0,1,0,-1},sum,n,m;
bool limit(int x,int y)//是否出界
{
	if(x>=0&&x<n&&y>=0&&y<m)
	return true;
	else
	return false;
}
void dfs(int x,int y)
{
	map[x][y]='#';//更新地板
	for(int i=0;i<4;i++)
	{
		int xx=x+dir[i][0];
		int yy=y+dir[i][1];
		if(limit(xx,yy)&&map[xx][yy]=='.')
		{
			sum++;
			dfs(xx,yy);
		}
	}
}
int main()
{
	while(scanf("%d %d",&m,&n)!=EOF)
	{
		if(n==0&&m==0)
		break;
		memset(map,0,sizeof(map));
		for(int i=0;i<n;i++)
		scanf("%s",map[i]);
		sum=0;
		for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
		if(map[i][j]=='@')
		{
			dfs(i,j);
			break;
		}
		printf("%d\n",sum+1);//起点也是，所以+1
	}
	return 0;
}