POJ 2386 Lake Counting

来源:http://poj.org/problem?id=2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total 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 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.

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.   開始理解成至少要两个"W"才算一块水域---还有方向，仅仅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。不须要回溯).... 

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;
}