hdu1312 Red and Black

I - Red and Black

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u

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

#include<stdio.h>

#include<queue>

#include<string.h>

#include<iostream>

#include<algorithm>

using namespace std;

int n,m;

char a[25][25];

int b[25][25];

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

struct node

{

    int x,y;

} w;

int sum=0;

void bfs(int x,int y)

{

    queue<node> q;

    int k,i,j;

    w.x=x;

    w.y=y;

    q.push(w);

    while(!q.empty())

    {

        node s=q.front();

        q.pop();

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

        {

            w.x=s.x+dir[i][0];

            w.y=s.y+dir[i][1];

            if(!b[w.x][w.y]&&a[w.x][w.y]!='#'&&w.x>=1&&w.x<=m&&w.y>=1&&w.y<=n)

            {

                sum++;

                b[w.x][w.y]=1;

                q.push(w);

            }

        }

    }

}

int main()

{

    while(~scanf("%d%d",&n,&m)&&(n||m))

    {

        sum=0;

        int i,j,x=0,y=0;

        memset(a,0,sizeof(a));

        memset(b,0,sizeof(b));

        for(i=1; i<=m; ++i)

            scanf("%s",a[i]+1);

        for(i=1; i<=m; ++i)

            for(j=1; j<=n; ++j)

            {

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

                {

                    x=i;

                    y=j;

                    break;

                }

            }

        b[x][y]=1;

        bfs(x,y);

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

    }

    return 0;

}

//#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;

//    }

//}

//

//

//

//#include <iostream>

//

//using namespace std;

//char a[50][50];

//int m,n;

//int sum;

//void dfs(int i,int j){

//    a[i][j]='#';//把搜素到的点变成#

//    sum++;//sum用于计数每搜到一个点就记一下数

//    if(i-1>=0&&a[i-1][j]=='.')dfs(i-1,j);

//    if(i+1<m&&a[i+1][j]=='.')dfs(i+1,j);

//    if(j-1>=0&&a[i][j-1]=='.')dfs(i,j-1);

//    if(j+1>=0&&a[i][j+1]=='.')dfs(i,j+1);//四个方向搜素

//}

//int main()

//{

//    while(cin>>n>>m&&(m+n)){

//        for(int i=0; i<m; i++)

//        cin>>a[i];

//        sum=0;

//        for(int i=0; i<m; i++)

//        for(int j=0; j<n; j++){

//            if(a[i][j]=='@')

//                dfs(i,j);

//        }

//        cout<<sum<<endl;

//    }

//    return 0;

//}

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hdu1312 Red and Black

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