Red and Black-HDU1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25113    Accepted Submission(s): 15166

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

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

题意:从@这个起点出发，求所有能走的.的个数，#为墙

思路:dfs写，用一个vis数组记录是否已经走过，

代码:

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
char k[1000][1000];
int n,m,sum,c1,c2;
int x[4]={-1,0,0,1};
int y[4]={0,-1,1,0};
int vis[1000][1000];
void dfs(int a,int b)
{
	for(int i=0;i<4;i++){
		int sx=x[i]+a;
		int sy=y[i]+b;
		if(sx<0||sx>=m||sy<0||sy>=n||vis[sx][sy]==1||k[sx][sy]=='#') continue;//sx,sy对应的n,m一定要注意 
		if(k[sx][sy]=='.'&&vis[sx][sy]==0){//vis判断是否已经访问 
			sum++;	
			vis[sx][sy]=1;		
			dfs(sx,sy);//不在此处在sum+1,回溯的时候会变？？ 
		}
	}
}
int main()
{
	while(~scanf("%d%d",&n,&m),n&&m){//若在此处添加\n，而不是在下方用getchar(); 
	    getchar();                   //则遇到0 0会无法退出 
		memset(k,'\0',sizeof(k));
		memset(vis,0,sizeof(vis)); 
		sum=1;
		for(int i=0;i<m;i++){
			for(int j=0;j<n;j++){
				cin>>k[i][j];
				if(k[i][j]=='@') c1=i,c2=j;
			}
		}
		dfs(c1,c2);
		printf("%d\n",sum);
	}
}

提供第二种写法的dfs，这个效率更高一点，写的也少

代码：

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
char k[1000][1000];
int n,m,sum,c1,c2;
int x[4]={-1,0,0,1};
int y[4]={0,-1,1,0};
void dfs(int a,int b)
{
	sum++;
	k[a][b]='#';
	for(int i=0;i<4;i++){
		int sx=x[i]+a;
		int sy=y[i]+b;
		if(sx>0&&sx<=n&&sy>0&&sy<=m&&k[sx][sy]=='.'){//若起始的i,j,是否是从1开始，注意边界判断 
			dfs(sx,sy);
		}
	}
}
int main()
{
	while(~scanf("%d%d",&m,&n),n&&m){//若在此处添加\n，而不是在下方用getchar(); 
	    getchar();                   //则遇到0 0会无法退出 
		memset(k,'\0',sizeof(k));
		sum=0;
		for(int i=1;i<=n;i++){
			for(int j=1;j<=m;j++){
				cin>>k[i][j];
				if(k[i][j]=='@') c1=i,c2=j;
			}
		}
		dfs(c1,c2);
		printf("%d\n",sum);
	}
}