• POJ2386:Lake Counting(DFS) java程序员


    Lake Counting
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 14776   Accepted: 7496

    Description

    Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors. 

    Given a diagram of Farmer John's field, determine how many ponds he has.

    Input

    * Line 1: Two space-separated integers: N and M 

    * Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

    Output

    * Line 1: The number of ponds in Farmer John's field.

    Sample Input

    10 12
    W........WW.
    .WWW.....WWW
    ....WW...WW.
    .........WW.
    .........W..
    ..W......W..
    .W.W.....WW.
    W.W.W.....W.
    .W.W......W.
    ..W.......W.

    Sample Output

    3

    Hint

    OUTPUT DETAILS: 

    There are three ponds: one in the upper left, one in the lower left,and one along the right side.

    Source

    MYCode:

    #include<iostream>
    #include<cstring>
    #include<cstdio>
    #define MAX 110
    char map[MAX][MAX];
    bool vis[MAX][MAX];
    int dirt[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
    using namespace std;
    void dfs(int x,int y)
    {
        vis[x][y]=1;
        int i;
        int tx,ty;
        for(i=0;i<8;i++)
        {
            tx=x+dirt[i][0];
            ty=y+dirt[i][1];
            if(!vis[tx][ty] && map[tx][ty]=='W')
            {
                dfs(tx,ty);
            }
        }
    }

    int main()
    {
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            int i,j;
            memset(vis,0,sizeof(vis));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    cin>>map[i][j];
                }
            }
            int ans=0;
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                {
                    if(map[i][j]=='W'&&!vis[i][j])
                    {
                        ans++;
                        dfs(i,j);
                    }
                }
            }
            printf("%d\n",ans);
        }
    }

    //

    DFS

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  • 原文地址:https://www.cnblogs.com/java20130725/p/3215882.html
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