HDU1312 Red and Black (DFS || BFS)

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

     
     
     
     

      
      
      
      45
59
6
13 
     
     
     
     

题意：从起点出发，能踩到的格子数。

dfs代码如下：

#include <iostream>
#include <cstring>
using namespace std;
char a[30][30];
int flag[30][30],sum;

void dfs(int i,int j)
{
	if(a[i-1][j]=='.'&&flag[i-1][j]==0)
	{
		sum++;
		a[i-1][j]='#';
		flag[i-1][j]=1;
		dfs(i-1,j);
	}
	if(a[i][j-1]=='.'&&flag[i][j-1]==0)
	{
		sum++;
		a[i][j-1]='#';
		flag[i][j-1]=1;
		dfs(i,j-1);
	}
	if(a[i][j+1]=='.'&&flag[i][j+1]==0)
	{
		sum++;
		a[i][j+1]='#';
		flag[i][j+1]=1;
		dfs(i,j+1);
	}
	if(a[i+1][j]=='.'&&flag[i+1][j]==0)
	{
		sum++;
		a[i+1][j]='#';
		flag[i+1][j]=1;
		dfs(i+1,j);
	}
	
}

int main()
{
	int w,h,i,j;
	while(cin>>w>>h,w+h)
	{
		memset(flag,0,sizeof(flag));
		memset(a,'#',sizeof(a));
		for(i=1;i<=h;i++)
		{
			for(j=1;j<=w;j++)
			{
				cin>>a[i][j];
			}
		}
//		for(i=0;i<w;i++)
//		{
//			for(j=0;j<h;j++)
//			{
//				cout<<a[i][j];
//			}
//		}
		sum=0;
		int x,y;
		for(i=1;i<=h;i++)
		{
			for(j=1;j<=w;j++)
			{
				if(a[i][j]=='@')
				{
					sum++;
					flag[i][j]=1;
					dfs(i,j);
					i=h;j=w;
				}
			}
		}
		cout<<sum<<endl;
	 } 
	 return 0;
}

bds代码如下：

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
typedef struct
{
    int x,y;
}node;
int n,m,a,b;
char c[25][25];
void sert()
{
    int i,f[4][2]={0,1,0,-1,1,0,-1,0},p=0;;
    queue<node> q;  
    node t,temp;
    t.x=a;
    t.y=b;
    q.push(t);
    while(!q.empty())
    {
       t=q.front();
       q.pop();
       for(i=0;i<4;i++)
       {
           temp.x=t.x+f[i][0];
           temp.y=t.y+f[i][1];
           if(temp.x>=0&&temp.x<m&&temp.y<n&&temp.y>=0&&c[temp.x][temp.y]!='#')
           {
               p++;
               q.push(temp);
               c[temp.x][temp.y]='#';
           }
       }
    }
    printf("%d\n",p);
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m),n||m)
    {
        for(i=0;i<m;i++)
        {
            getchar();
            for(j=0;j<n;j++)
            {
                scanf("%c",&c[i][j]);
                if(c[i][j]=='@')
                {
                    a=i;
                    b=j;
                }
            }
        }
            sert();
    }
    return 0;
}
//bfs 转载http://www.cnblogs.com/xiaofanke/archive/2012/08/22/2651350.html 

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