poj2386-lak counting

最新推荐文章于 2019-04-27 20:27:27 发布

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本文介绍了一个湖泊计数问题的解决方案，通过深度优先搜索（DFS）遍历地图来确定由水形成的独立湖泊数量。该问题源自USACO 2004年11月的比赛，涉及在一个由N*M个格子组成的矩形区域内寻找彼此相连的水池。

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Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36990		Accepted: 18396

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

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

题目意思八连通的积水是连在一起的，所以，从任意的'W'开始，不停的把邻接的用'.'代替，一次dfs后，所有与'W'联通的W就被替换成'.'，总共进行的dfs的次数就是结果

#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn 101
using namespace std;
char map[maxn][maxn];
int N, M;
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;
			int ny = y + dy;
			if (0 <= nx&&nx < N && 0 <= ny&&ny < M&&map[nx][ny] == 'W') dfs(nx, ny);
		}
	}
	return ;
}
int main() {
	scanf("%d %d", &N, &M);
	getchar();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			scanf("%c",&map[i][j]);
		}
		getchar();
	}
	int cnt = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			if (map[i][j] == 'W') {
				dfs(i, j);
				cnt++;
			}
		}
	}
	printf("%d\n", cnt);
	return 0;
}