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
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* 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
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3
- 提示
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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 }
