Poj2386 Lake Counting （DFS）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 49414		Accepted: 24273

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.

 #include <iostream>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 char a[][];
 int n,m;
 void dfs(int x,int y)
 {
     a[x][y]='.';
     for(int dx=-;dx<=;dx++){
         for(int dy=-;dy<=;dy++){
             int nx=x+dx,ny=y+dy;
             if(nx>=&&nx<n&&ny>=&&ny<m&&a[nx][ny]=='W'){
                 dfs(nx,ny);
             }
         }
     }
     return ;
 }
 void solve()
 {
     int res=;
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(a[i][j]=='W'){
                 dfs(i,j);
                 res++;
             }
         }
     }
     cout<<res<<endl;
 }
 int main()
 {
     while(cin>>n>>m){
         for(int i=;i<n;i++){
             for(int j=;j<m;j++){
                 cin>>a[i][j];
             }
         }
         solve();
     }
     return ;
 }

自己再熟悉一遍：

 #include <iostream>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <cstring>
 using namespace std;
 int n,m;
 char a[][];
 int res;
 int nx,ny;
 void dfs(int x,int y)
 {
     //八个方向搜索
     for(int i=-;i<=;i++){
         for(int j=-;j<=;j++){
             nx=x+i,ny=y+j;
             if(nx>=&&nx<n&&ny>=&&ny<m&&a[nx][ny]=='W'){
                 a[nx][ny]='.';
                 dfs(nx,ny);
             }
         }
     }
 }
 void solve()
 {
     res=;
     //每次从发现W的时候开始搜索由于一簇只能算一个地方所以res计数只加一次
     for(int i=;i<n;i++){
         for(int j=;j<m;j++){
             if(a[i][j]=='W'){
                 res++;
                 dfs(i,j);
             }
         }
     }
 }
 int main()
 {
     while(cin>>n>>m){
         memset(a,,sizeof(a));
         for(int i=;i<n;i++){
             for(int j=;j<m;j++){
                 cin>>a[i][j];
             }
         }
         solve();
         cout<<res<<endl;
     }
     return ;
 }