1388:Lake Counting

题目链接：

NOI题库http://noi.openjudge.cn/ch0205/1388/

POJ 2386 http://poj.org/problem?id=2386

总时间限制: 1000ms  内存限制: 65536kB

描述

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.

输入

* 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.

输出

* Line 1: The number of ponds in Farmer John's field.

样例输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出

3

提示

OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.

题目大意：

有一块N*M的土地，雨后机起了水，有水标记为'W'，干燥标记为'.'。
八连通的积水被认为是连接在一起的。需要求出院子里有多少水洼。
首先输入N和M，然后输入N*M的二维字符数组，行内字符之间没有空格。
输出水洼的数目。
N和M都在100以内。

算法分析:
这道题可以用深搜，也可以用广搜。

扫描二维数组，每遇到一个'W'就从这个地方出发开始深搜或广搜。
搜索过程中，把搜缩遇到的'W'全部设为'.'。
每当完成一次深搜或广搜，水洼数增1.

题解可以参考：
http://blog.csdn.net/c20180630/article/details/52329915
http://www.cnblogs.com/sooner/archive/2013/04/09/3010938.html

深搜的思路：

下面是我自己写的代码，含深搜与广搜的代码：

 #include<stdio.h>
 #include<iostream>
 #include<queue>
 #include<stack>
 using namespace std;

 struct obj
 {
     int xx,yy;
 };

 int N,M;
 char a[][];
 int Count;

 int dx[]={-,-,,,,,,-};//从正上方开始，顺时针旋转的8个方向
 int dy[]={,,,,,-,-,-};
 void BFS(int x,int y);//从(x,y)开始广搜
 void DFS(int x,int y);//从(x,y)开始深搜
 void DFS2(int x,int y);//从(x,y)开始深搜，递归实现 

 int main(int argc, char *argv[])
 {
     freopen("1388.in","r",stdin);
     int i,j;
     scanf("%d%d",&N,&M);getchar();//吸收回车符
     for(i=;i<N;i++)
     {
         gets(a[i]);
         /*
         for(j=0;j<M;j++)
         {
             a[i][j]=getchar();
         }
         getchar();//吸收回车符
         */
     }

     Count=;
     for(i=;i<N;i++)
     {
         for(j=;j<M;j++)
         {
             if(a[i][j]=='W')
             {
                 BFS(i,j);//从（i，j）开始广搜
                 //DFS(i,j);//从（i，j）开始深搜
                 //{ DFS2(i,j); Count=Count+1;} //递归实现的深搜，从(i,j)开始深搜
             }
         }
     }
     printf("%d\n",Count);
     return ;
 }
 void BFS(int x,int y)//从（x,y）开始广搜
 {
     queue<struct obj> q;
     struct obj start,temp;
     int i,txx,tyy;

     start.xx=x;
     start.yy=y;
     q.push(start);
     a[x][y]='.';
     while(!q.empty())
     {
         for(i=;i<;i++)
         {
             txx=q.front().xx+dx[i];
             tyy=q.front().yy+dy[i];
             if(txx>=&&txx<N&&tyy>=&&tyy<M&&a[txx][tyy]=='W')
             {
                 temp.xx=txx;
                 temp.yy=tyy;
                 a[txx][tyy]='.';
                 q.push(temp);
             }
         }
         q.pop();
     }
     Count++;
 }

 void DFS(int x,int y)//从(x,y)开始深搜
 {
     stack<struct obj> s;
     struct obj start,temp,temp2;
     int i,txx,tyy;

     a[x][y]='.';
     start.xx=x;
     start.yy=y;
     s.push(start);
     while(!s.empty())
     {
         temp=s.top();  s.pop();
         for(i=;i<;i++)
         {
             txx=temp.xx+dx[i];
             tyy=temp.yy+dy[i];
             if(txx>=&&txx<N&&tyy>=&&tyy<M&&a[txx][tyy]=='W')
             {
                 temp2.xx=txx;
                 temp2.yy=tyy;
                 a[txx][tyy]='.';
                 s.push(temp2);
             }
         }
     }
     Count++;
 }

 void DFS2(int x,int y)//从(x,y)开始深搜，递归实现
 {
     int i,txx,tyy;
     a[x][y]='.';
     for(i=;i<;i++)
     {
         txx=x+dx[i];
         tyy=y+dy[i];
         if(txx>=&&txx<N&&tyy>=&&tyy<M&&a[txx][tyy]=='W')
         {
             //a[txx][tyy]='.';
             DFS2(txx,tyy);
         }
     }
 }