Red and Black 搜索

转载自http://blog.csdn.net/zone_programming/article/details/6688424  DFS

转载自http://blog.163.com/cristiano_faith/blog/static/1725054812011116103645188/ BFS

Red and Black

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 16   Accepted Submission(s) : 11

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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

45
59
6
13

Source

Asia 2004, Ehime (Japan), Japan Domestic

搜索真心不会：来自转载：

   
   
   
   
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int direct[4][2] = { -1, 0, 1, 0, 0, 1, 0, -1 };  /*定义方向， 左右，上下*/

char map[21][21];               /*输入的字符串*/
bool mark[21][21];              /*标记走过的路程*/
bool flag;
int W, H;
int Dx, Dy;            //记录起始位置@，从这里开始进行搜索
int ans;                //记录满足的个数。初始化为1，因为@也包含在内
/*****底下是核心算法，主要是从
上下左右这四个方向进行搜索，注意
满足搜索的条件是不能越界，不能是#，
还有就是没有搜索过的--》主要是靠mark[i][j]
来实现*******/
void DFS( int x, int y )
{
    mark[x][y] = true;
    for( int k = 0; k < 4; k ++ )
    {
        int p = x + direct[k][0];
        int q = y + direct[k][1];
        if( p >= 0 && q >= 0 && p < H && q < W && !mark[p][q] && map[p][q] != '#' )
        {
            ans ++;
            DFS( p, q );
        }
    }
    return;
}

int main()
{
    int i, j, k;
    while( cin >> W >> H && ( W || H ) )   // W -> column, H -> row;
    {
        memset( mark, false, sizeof( mark ) );
        for( i = 0; i < H; i ++ )
            for( j = 0; j < W; j ++ )
            {
                cin >> map[i][j];
                if( map[i][j] == '@' )
                {
                    Dx = i;
                    Dy = j;
                }
            }
        ans = 1;
        DFS( Dx, Dy );
        cout << ans << endl;
    }
}

题目大意：一个矩形空间，上有黑色同红色两种砖，求从某一黑砖出发可以经过的黑砖的数目

 原先用DFS做的，但调试很久都不知哪里出错，后来改用BFS  

#include<stdio.h>

#include<queue>

using namespace std;

int n,m,total;

int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

char map[20][20];

typedef struct

{

    int x,y;

}point;

point start;

int bfs(point start)

{

    queue<point>q;

    int i;

    point cur,next;

    map[start.x][start.y]='#';

    q.push(start);

    while(!q.empty())

    {

       cur=q.front();

       q.pop();

       for(i=0;i<4;i++)

       {

           next.x=cur.x+dir[i][0];

           next.y=cur.y+dir[i][1];

           if(next.x>=0&&next.x<m&&next.y>=0&&next.y<n)

              if(map[next.x][next.y]!='#')

              {

                  map[next.x][next.y]='#';

                  total++;

                  q.push(next);

              }

 

       }

    }

    return total;

   

}

int main()

{

    int i,j;

    while(scanf("%d%d",&n,&m)!=EOF&&n!=0)

    {  

       getchar();

       for(i=0;i<m;i++)

       {

           for(j=0;j<n;j++)

           {  

              scanf("%c",&map[i][j]);

              if(map[i][j]=='@')

              {start.x=i;start.y=j;}

           }

           getchar();

       }

       total=1;

       printf("%d\n",bfs(start));

    }

    return 0;

}