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

BFS

#include<cstdio>

#include<iostream>

#include<cstring>

using namespace std;

int w,h,sx,sy,cnt;

char map[30][30];

int  vis[30][30];

int dx[]={0,0,-1,1};

int dy[]={1,-1,0,0};

struct Node

{

    int x;

    int y;

}Q[450];

Node s;

void bfs()

{

    int front=0,rear=0;

    Q[rear++]=s;

    while(front<rear)

    {

        Node t=Q[front++];

        for(int i=0;i<4;i++)

    {

        int x0=t.x+dx[i];

        int y0=t.y+dy[i];

        Node f;

        f.x=x0;f.y=y0;

        if(!vis[f.x][f.y]&&f.x>=0&&f.x<h&&f.y>=0&&f.y<w&&map[f.x][f.y]!='#')

        {

            vis[f.x][f.y]=1;

            Q[rear++]=f;

            if(map[f.x][f.y]=='.')

             cnt++;

        }

    }

    }

}

int main()

{

    while(~scanf("%d%d",&w,&h))

    {if(w==0||h==0)

    break;

        memset(vis,0,sizeof(vis));

        memset(map,0,sizeof(map));

      for(int i=0;i<h;i++)

        {

        scanf("%s",map[i]);

        for(int j=0;j<w;j++)

            if(map[i][j]=='@')

            {

                s.x=i;

                s.y=j;

                break;

            }

        }

        cnt=0;

        bfs();

        printf("%d\n",cnt+1);}

    return 0;

}

DFS

#include<cstdio>

#include<iostream>

#include<cstring>

using namespace std;

int w,h,sx,sy,cnt;

char map[30][30];

int  vis[30][30];

int dx[]={0,0,-1,1};

int dy[]={1,-1,0,0};

void dfs(int x,int y)

{

    if(map[x][y]=='.')

    cnt++;

    if(x<0||x>=h||y<0||y>=w||map[x][y]=='#')

    return;

    for(int i=0;i<4;i++)

    {

        int x0=x+dx[i];

        int y0=y+dy[i];

        if(!vis[x0][y0])

        {

            vis[x0][y0]=1;

            dfs(x0,y0);

        }

    }

}

int main()

{

    while(~scanf("%d%d",&w,&h))

    {if(w==0||h==0)

    break;

        memset(vis,0,sizeof(vis));

        memset(map,0,sizeof(map));

      for(int i=0;i<h;i++)

        {

        scanf("%s",map[i]);

        for(int j=0;j<w;j++)

            if(map[i][j]=='@')

            {

                sx=i;

                sy=j;

                break;

            }

        }

        cnt=0;

        dfs(sx,sy);

        printf("%d\n",cnt+1);}

    return 0;

}

//6 9

//....#.

//.....#

//......

//......

//......

//......

//......

//#@...#

//.#..#.

可以看出：写BFS时一般要有结构体来表示状态。

求最短路一般用BFS，其他的可能更多用的是DFS

两者的关键都在于找转态。

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