Lake Counting

Lake Counting

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

 
 
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Source Code:
#include <iostream>

using namespace std;

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

const int arrSize=110;
char maze[arrSize][arrSize];

void DFS(int x,int y,int n,int m){
    maze[x][y]='.';
    for(int i=0;i<8;++i){
        int newX=x+go[i][0];
        int newY=y+go[i][1];
        if(newX<0||newX>=n)
            continue;
        if(newY<0||newY>=m)
            continue;
        if(maze[newX][newY]=='.')
            continue;
        DFS(newX,newY,n,m);
    }
    return;
}

int main()
{
    int N,M;
    int pondsCount;
    while(cin>>N>>M){
        for(int i=0;i<N;++i)
            for(int j=0;j<M;++j)
                cin>>maze[i][j];
        pondsCount=0;
        for(int i=0;i<N;++i){
            for(int j=0;j<M;++j){
                if(maze[i][j]=='W'){
                    DFS(i,j,N,M);
                    ++pondsCount;
                }
            }
        }
        cout<<pondsCount<<endl;
    }
    return 0;
}

 

 
posted @ 2015-01-11 22:00  tinylcy  阅读(202)  评论(0编辑  收藏  举报