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Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 34073 Accepted: 16954

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 12W........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

USACO 2004 November

从小白做起，很痛苦，做一只永远拖不跨，打不倒的小强

#include <iostream>
#include <cstring>

using namespace std;

char a[109][109];
int step[8][2]={{-1,-1},{0,-1},{1,-1},{-1,0},{1,0},{-1,1},{0,1},{1,1}};

void dfs(int x,int y)
{
    for (int i=0;i<8;i++)
    {
        int A=x+step[i][0];
        int B=y+step[i][1];
        if (a[A][B]==-1||a[A][B]=='.')
        {
            continue;
        }
        else
        {
            a[A][B]='.';
            dfs(A,B);
        }
    }
}
int main()
{
    int N,M;
    cin>>N>>M;
    memset(a,-1,sizeof(a));
    for (int i=1;i<=N;i++)
    {
        for (int j=1;j<=M;j++)
        {
            cin>>a[i][j];
        }
    }
    int sum=0;
    for (int i=1;i<=N;i++)
    {
        for (int j=1;j<=M;j++)
        {
            if (a[i][j]=='W')
            {
                //cout<<i<<' '<<j<<endl;
                //a[i][j]='.';
                sum++;
                dfs(i,j);
            }
        }
    }
    cout<<sum;
    return 0;
}