POJ 2386 Lake Counting

链接：http://poj.org/problem?id=2386

Lake Counting

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 24263 Accepted: 12246

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

大意——给你一个n*m的矩形网格，每个格子里面要么是‘W’，要么是‘.’，它们分别表示水和旱地。现在要你计算出有多少个池塘。每个池塘由若干个水组成，水能连接的方向有8个，只要是能连接到的水都属于一个池塘。

思路——一个简单的DFS题。我们将所有的网格全部遍历一次，假如我们找到一个池塘的源头，就可以进行一次计数，并且将目前找到的池塘源头标记为‘.’，那么下一次就不会重复访问了，再从八个方向进行深搜，将能到达相邻的‘W’全部标记，以免重复访问。这样最后得到的计数就是答案。

复杂度分析——时间复杂度:O(n*m),空间复杂度:O(n*m)

附上AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef unsigned int li;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxn = 105;
const int dir[8][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {1, -1}, {-1, 1}, {1, 1}}; // 8个方向
int n, m; // 矩形的行列
char mat[maxn][maxn]; // 矩形

void dfs(int x, int y); // 深度优先搜索

int main()
{
	ios::sync_with_stdio(false);
	while (~scanf("%d%d", &n, &m))
	{
		int cnt = 0;
		for (int i=0; i<n; i++)
			scanf("%s", mat[i]);
		for (int i=0; i<n; i++)
			for (int j=0; j<m; j++)
				if (mat[i][j] == 'W')
				{ // 找到池塘源头，计数并深搜
					cnt++;
					dfs(i, j);
				}
		printf("%d\n", cnt);
	}
	return 0;
}

void dfs(int x, int y)
{
	mat[x][y] = '.'; // 访问过了，标记
	for (int i=0; i<8; ++i) // 从八个方向找相邻的
		if (x+dir[i][0]>=0 && x+dir[i][0]<n && y+dir[i][1]>=0 &&
			y+dir[i][1]<m && mat[x+dir[i][0]][y+dir[i][1]]=='W')
			dfs(x+dir[i][0], y+dir[i][1]); // 找到相邻的，继续深搜
}