poj 2386--Lake Counting

Lake Counting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 37283 Accepted: 18532
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

USACO 2004 November

解题思路：
采用dfs（深度优先搜索）的方法，先由上到下，由左至右的搜索当搜索到 ‘W’ 时进行dfs，记录下dfs的次数（即为结果）。
在dfs时将此处的 ‘W’ 改为 ‘.’,再进行下一次dfs，以此递归。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>typedef long long LL;using namespace std;const int MAX = 1E2+10;char field[MAX][MAX]; int N,M;void dfs(int x,int y){    field[x][y]='.';//将此处'W'改为'.',表示已经搜索过了；    for(int dx=-1;dx<=1;dx++)        for(int dy=-1;dy<=1;dy++)        {            int xx=x+dx,yy=y+dy;//向四周进行搜索；            if(xx>=0 && xx<N && yy>=0 && yy<M && field[xx][yy]=='W') dfs(xx,yy);//递归到下一次dfs；        }}int main(){    int t=0;    cin>>N>>M;    for(int i=0;i<N;i++)        scanf("%s",field[i]);    for(int i=0;i<N;i++)    {        for(int j=0;j<M;j++)        {            if(field[i][j]=='W')            {                t++;//记录下dfs的次数，即为结果；                dfs(i,j);            }        }     }     cout<<t<<endl;    return 0;}