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  • Lake Counting

    题号:POJ 2386

    题目:

    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.
     
     
    分析:题目给出一个地图,W代表水坑,.代表干地,要你求图中共有多少个大水坑(连在一起的小水坑算一个,周围的九个方位都算),差不多就是一个DFS的模板题,从找的第一个W出发,寻找周围有的‘W',找到就让它为’ .',不然会重复操作;
     
     
     
    AC代码:
    #include<iostream>
    #include<cstdio>
    #define N 125
    using namespace std;
    char a[N][N];
    int row,col;
    void dfs(int x,int y)
    {
        a[x][y]='.';
        for (int dx=-1;dx<=1;dx++)
          for (int dy=-1;dy<=1;dy++)
        {
            int nx=x+dx,ny=y+dy;
            if (nx>=0&&nx<=row&&ny>=0&&ny<=col&&a[nx][ny]=='W')
                dfs(nx,ny);
        }
    }
    int main()
    {
    
        cin>> row>> col;
        for (int i = 0; i <row; i++)
            for (int j = 0; j <col; j++)
                cin >>a[i][j];
        int res = 0;
        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
                if (a[i][j] == 'W')//从有积水的地方开始dfs
                {
                    dfs(i, j);
                    res++;
                }
        cout << res << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/lisijie/p/7739146.html
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