POJ 2386 Lake Counting

来源:http://poj.org/problem?id=2386

Lake Counting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20124 Accepted: 10139

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 12W........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

题意:题意题意！！！ 一句话两处理解错误,,,,,    求一块田野里有多少块水域～～ 里面最关键的一句：A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.  开始理解成至少要两个"W"才算一块水域---还有方向，只考虑上下左右四个方向，导致一直WA..

题解: DFS(感觉又不是标准的DFS，不需要回溯).... 

AC代码:

#include<iostream>#include<string>using namespace std;int dir[8][2]={    {0,1},{0,-1},    {1,0},{-1,0},    {1,1},{1,-1},    {-1,-1},{-1,1}};string map[105];int dx,dy,count=0;bool flag;void dfs(int x,int y){    for(int i=0;i<8;i++){    int tempx=x+dir[i][0],tempy=y+dir[i][1];    if(tempx>=0&&tempx<dx&&tempy>=0&&tempy<dy&&map[tempx][tempy]=='W'){    map[tempx][tempy]='.';    dfs(tempx,tempy);    }    }}int main(){    cin>>dx>>dy;    for(int i=0;i<dx;i++)    cin>>map[i];    for(int i=0;i<dx;i++)    for(int j=0;j<dy;j++){    if(map[i][j]=='W'){    count++;    map[i][j]='.';    dfs(i,j);    }    }    cout<<count<<endl;    return 0;}