POJ - 2386 Lake Counting

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 44131		Accepted: 21800

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：

由于近日阴雨连天，约翰的农场中中积水汇聚成一个个不同的池塘，农场可以用 N x M (1 <= N <= 100; 1 <= M <= 100) 的矩形来表示。农场中的每个格子可以用'W'或者是'.'来分别代表积水或者土地，约翰想知道他的农场中有多少池塘。八连通的积水被认为是一块积水。 

给你约翰农场的航拍图，确定有多少池塘？

Input

* 第1行：N 和 M 

*第2行：N+1: M个字符一行，每个字符代表约翰的农场的土地情况。每个字符中间不包含空格。

Output

* 第1行：池塘的数量

题解：用深度搜索dfs，联通的只是一块，要把这一块都标记后，才能搜索下一块。

深度搜索dfs侧重于解决（求出符合题目要求的所有解法个数 ）的问题。

与深度搜索dfs不同的是，广搜bfs旨在于找出最短的路径，最好的结果。

#include<iostream>

using namespace std;
const int MAX = 110;
int N,M;
char Map[MAX][MAX];
void dfs(int x,int y){
    Map[x][y] = '.';
    for(int dx = -1;dx <= 1;++dx){
        for(int dy = -1;dy <= 1;++dy){
            int nx = x + dx,ny = y + dy;
            if(1 <= nx && nx <= N && 1 <= ny && ny <= M){
                if(Map[nx][ny] == 'W'){
                    dfs(nx,ny);
                }
            }
        }
    }
}
int solve(){
    int res = 0;
    for(int i=1;i<=N;++i){
        for(int j=1;j<=M;++j){
            if(Map[i][j] == 'W'){
                dfs(i,j);
                res++;
            }
        }
    }
    return res;
}
int main(void){
    cin >> N >> M;
    for(int i=1;i<=N;++i){
        for(int j=1;j<=M;++j){
            cin >> Map[i][j];
        }
    }
    cout << solve() << endl;
    return 0;
}