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：

  1 #include<stdio.h>
  2 #include<iostream>
  3 #include<queue>
  4 #include<stack>
  5 using namespace std;
  6
  7 struct obj
  8 {
  9     int xx,yy;
 10 };
 11
 12 int N,M;
 13 char a[102][102];
 14 int Count;
 15
 16 int dx[8]={-1,-1,0,1,1,1,0,-1};//从正上方开始，顺时针旋转的8个方向
 17 int dy[8]={0,1,1,1,0,-1,-1,-1};
 18 void BFS(int x,int y);//从(x,y)开始广搜
 19 void DFS(int x,int y);//从(x,y)开始深搜
 20 void DFS2(int x,int y);//从(x,y)开始深搜，递归实现
 21
 22 int main(int argc, char *argv[])
 23 {
 24     freopen("1388.in","r",stdin);
 25     int i,j;
 26     scanf("%d%d",&N,&M);getchar();//吸收回车符
 27     for(i=0;i<N;i++)
 28     {
 29         gets(a[i]);
 30         /*
 31         for(j=0;j<M;j++)
 32         {
 33             a[i][j]=getchar();
 34         }
 35         getchar();//吸收回车符
 36         */
 37     }
 38
 39     Count=0;
 40     for(i=0;i<N;i++)
 41     {
 42         for(j=0;j<M;j++)
 43         {
 44             if(a[i][j]==‘W‘)
 45             {
 46                 BFS(i,j);//从（i，j）开始广搜
 47                 //DFS(i,j);//从（i，j）开始深搜
 48                 //{ DFS2(i,j); Count=Count+1;} //递归实现的深搜，从(i,j)开始深搜
 49             }
 50         }
 51     }
 52     printf("%d\n",Count);
 53     return 0;
 54 }
 55 void BFS(int x,int y)//从（x,y）开始广搜
 56 {
 57     queue<struct obj> q;
 58     struct obj start,temp;
 59     int i,txx,tyy;
 60
 61     start.xx=x;
 62     start.yy=y;
 63     q.push(start);
 64     a[x][y]=‘.‘;
 65     while(!q.empty())
 66     {
 67         for(i=0;i<8;i++)
 68         {
 69             txx=q.front().xx+dx[i];
 70             tyy=q.front().yy+dy[i];
 71             if(txx>=0&&txx<N&&tyy>=0&&tyy<M&&a[txx][tyy]==‘W‘)
 72             {
 73                 temp.xx=txx;
 74                 temp.yy=tyy;
 75                 a[txx][tyy]=‘.‘;
 76                 q.push(temp);
 77             }
 78         }
 79         q.pop();
 80     }
 81     Count++;
 82 }
 83
 84 void DFS(int x,int y)//从(x,y)开始深搜
 85 {
 86     stack<struct obj> s;
 87     struct obj start,temp,temp2;
 88     int i,txx,tyy;
 89
 90     a[x][y]=‘.‘;
 91     start.xx=x;
 92     start.yy=y;
 93     s.push(start);
 94     while(!s.empty())
 95     {
 96         temp=s.top();  s.pop();
 97         for(i=0;i<8;i++)
 98         {
 99             txx=temp.xx+dx[i];
100             tyy=temp.yy+dy[i];
101             if(txx>=0&&txx<N&&tyy>=0&&tyy<M&&a[txx][tyy]==‘W‘)
102             {
103                 temp2.xx=txx;
104                 temp2.yy=tyy;
105                 a[txx][tyy]=‘.‘;
106                 s.push(temp2);
107             }
108         }
109     }
110     Count++;
111 }
112
113 void DFS2(int x,int y)//从(x,y)开始深搜，递归实现
114 {
115     int i,txx,tyy;
116     a[x][y]=‘.‘;
117     for(i=0;i<8;i++)
118     {
119         txx=x+dx[i];
120         tyy=y+dy[i];
121         if(txx>=0&&txx<N&&tyy>=0&&tyy<M&&a[txx][tyy]==‘W‘)
122         {
123             //a[txx][tyy]=‘.‘;
124             DFS2(txx,tyy);
125         }
126     }
127 }