Red and Black(hdu-1312 搜索）

题目链接
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

1深搜：通过递归方法，从图结构的一个结点开始深度搜索。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int w,h,sum;
char mp[25][25];
int ds1[4]={-1,1,0,0};
int ds2[4]={0,0,-1,1};
void dfs(int x,int y)
{
    mp[x][y]='#';
    sum++;
    for(int i=0;i<4;i++)
    {
        int tx=x+ds1[i];//朝4个方向走
        int ty=y+ds2[i];
        if(tx>=0&&tx<h&&ty>=0&&ty<w&&mp[tx][ty]=='.')
        {
            dfs(tx,ty);
        }
    }
}

int main()
{
	while(cin>>w>>h)
	{
		if(w==0&&h==0)
			break;
		sum=0;
		int x,y;
		for(int i=0;i<h;i++)
			for(int j=0;j<w;j++)
				cin>>mp[i][j];
		for(int i=0;i<h;i++)
		{
			for(int j=0;j<w;j++)
			{
				if(mp[i][j]=='@')
				{
					x=i;
					y=j;
				}
			}
		}
		dfs(x,y);
		cout<<sum<<endl;
	}
	return 0;
}

2广搜：

#include<stdio.h>
#include<iostream>
#include<queue>
using namespace std;
typedef struct
{
    int x,y;
}node;
int w,h,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);//将t压入队列尾部
    //队列在队尾进行插入，在队头进行删除。
    while(!q.empty())
    {
       t=q.front();//队首元素的返回值赋给t
       q.pop();//删除队首元素
       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<h&&temp.y<w&&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",&w,&h))
    {
        if(w==0&&h==0)
            break;
        for(i=0;i<h;i++)
        {
            getchar();
            for(j=0;j<w;j++)
            {
                scanf("%c",&c[i][j]);
                if(c[i][j]=='@')
                {
                    a=i;
                    b=j;
                }
            }
        }
            sert();
    }
    return 0;
}