广度优先搜素和深度优先搜素

以北大的1979为例：

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 17144		Accepted: 9025

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

具体要求：

输入相应的格子的数目，并且标记每个格子的颜色，当遇到黑色的格子时可以踩上，当遇到红色的格子时不能踩上，要求输出从一个起始点出发能够踩到的黑色格子数。

算法思想：

1，可以根据图的深度优先搜索，计算并输出通过深度搜索所通过的节点数，即相应的格子数目。

2，可以根据图的广度优先搜索，计算并输出通过广度搜索所通过的节点数，即相应的格子数目。

算法实现：

1，深度优先搜索：通过递归方法，从图结构的一个结点开始深度搜索。相应的代码如下

#include<stdio.h>
#include<string.h>
int a,b,n,m;
char c[22][22];
int sert(int a,int b)
{
    if(c[a][b]=='#'||a>=m||b>=n||a<0||b<0)
        return 0;
    else
    {
        c[a][b]='#';
        return 1+sert(a,b+1)+sert(a,b-1)+sert(a-1,b)+sert(a+1,b);
    }
}
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;
                }
            }
        }
            n=sert(a,b);
            printf("%d\n",n);
    }
    return 0;
}

广度优先搜索：通过建立队列实现广度搜索，每走到符合要求的格子时，当前格子进队列。相应的代码如下：

#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;
}