[poj2349 Arctic Network]最小生成树-优快云博客

本文介绍了一种使用Kruskal算法来解决最小生成树问题的方法，目的是确定连接多个北部哨所所需的无线电发射器最小覆盖范围。文章详细解释了如何通过卫星通道和无线电连接各个哨所，并给出了解决方案的具体实现。

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10805		Accepted: 3542

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

Sample Output

212.13

Source

Waterloo local 2002.09.28
Kruskal搞的最小生成树。

Memory: 2676K		Time: 110MS
Language: G++		Result: Accepted

/*
	初始化每个点一个连通分量 
	每次只需要选最小边即可，然后在剩下s个连通分量时答案就更新完毕鸟
*/ 
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const double INF = 1e10;
const int MAX = 512;
struct Point {
	double x, y;
	double dis(const Point& B) {
		return sqrt((x - B.x) * (x - B.x) + (y - B.y ) * (y - B.y));
	}
} a[MAX];
struct edge {
	int from, to;
	double cost;
};
struct cmp {
	bool operator()(edge& e1, edge& e2) {
		return e1.cost > e2.cost;
	}
};

int father[MAX];

int find(int x) {
	return x == father[x] ? x : father[x] = find(father[x]);
}
void Union(int x, int y) {
	int fx = find(x), fy = find(y);
	if (fx != fy) {
		father[fx] = fy;
	}
}

double gao(int s, int p) {
	priority_queue<edge, vector<edge>, cmp> Q;
	for (int i = 0; i < p; ++i) {
		for (int j = i + 1; j < p; ++j) {
			Q.push((struct edge){i, j, a[i].dis(a[j])});
		}
		father[i] = i;
	}
	double ans = 0.0;
	for (int i = s; i < p; ++i) {
		edge e;
		e.from = e.to = e.cost = 0.0;
		while (!Q.empty()) {
			e = Q.top();
			Q.pop();
			if (find(e.from) != find(e.to)) break;
		}
		
		Union(e.from, e.to);
		if (ans < e.cost) ans = e.cost;
	}
	return ans;
}

int main() {
	int T;
	scanf(" %d", &T);
	while (T--) {
		int s, p;
		scanf(" %d %d", &s, &p);
		for (int i = 0; i < p; ++i) {
			scanf(" %lf %lf", &a[i].x, &a[i].y);
		}
		printf("%.2f\n", gao(s, p));
	}
	return 0;
}