POJ2386 Lake Counting

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本文介绍了一种通过深度优先搜索(DFS)算法来计算地图上形成的池塘数量的方法。地图由若干个水或陆地组成的网格构成，若水区域相连则视为同一池塘。文章提供了一个具体的C语言实现示例。

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8579		Accepted: 4367

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

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

【题意简述】

给出一个M*N的地图，有些地方被雨淹了。

问这些被淹没的地方形成了多少个池塘（连同块）？

( M,N<=100，这里规定任意格子周围的八个格子都是与它相邻的 )

用dfs+染色来实现：

其实和JOJ那道OIL 是一模一样的

#include<stdio.h>
char str[101][101];
int x[8]={-1,0,1,-1,1,-1,0,1},y[8]={-1,-1,-1,0,0,1,1,1};
int num,m,n;
void dfs(int x1,int y1)
{
     int i;
     str[x1][y1]='.';
     for(i=0;i<8;i++)
     if(x1+x[i]<m && x1+x[i]>=0 && y1+y[i]<n && y1+y[i]>=0 && str[x1+x[i]][y1+y[i]]=='W')
           dfs(x1+x[i],y1+y[i]);
}
int main()
{
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        num=0;
        for(int i=0;i<m;i++)
                 scanf("%s",str[i]);
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
              if(str[i][j]=='W')
                {
                     dfs(i,j);
                     num++;
                }
             printf("%d/n",num);
    }
    return 0;
}