UVALive 4872 Underground Cables 最小生成树

题目链接：

题目

Underground Cables
Time Limit: 3000MS
Memory Limit: Unknown
64bit IO Format: %lld & %llu

问题描述

A city wants to get rid of their unsightly power poles by moving their power cables underground. They have a list of points that all need to be connected, but they have some limitations. Their tunneling equipment can only move in straight lines between points. They only have room for one underground cable at any location except at the given points, so no two cables can cross.
Given a list of points, what is the least amount of cable necessary to make sure that every pair of points is connected, either directly, or indirectly through other points?

输入

There will be several test cases in the input. Each test case will begin with an integer N(2$ le$N$ le$1, 000), which is the number of points in the city. On each of the next N lines will be two integers, X and Y(- 1, 000$ le$X, Y$ le$1, 000), which are the (X, Y) locations of the N points. Within a test case, all points will be distinct. The input will end with a line with a single 0.

输出

For each test case, output a single real number, representing the least amount of cable the city will need to connect all of its points. Print this number with exactly two decimal places, rounded. Print each number on its own line with no spaces. Do not print any blank lines between answers.

样例

input
4
0 0
0 10
10 0
10 10
2
0 0
10 10
0

output
30.00
14.14

题意

给你n个点，求最少的线缆使得所有的点连在一起

题解

假设存在两根线交叉，那么明显存在一个不交叉的方案使这四个点连通，并且线缆总长度还要更小，所有我们构建完全图跑一遍最短生成树，是可以保证不会出现交叉边的。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;

const int maxn = 1010;
int n;

struct Point {
	int x, y;
}pt[maxn];

struct Edge {
	int u, v;
	double w;
	Edge(int u, int v, double w) :u(u), v(v), w(w) {}
	Edge() {}
	bool operator < (const Edge& e) {
		return w < e.w;
	}
}egs[maxn*maxn];

double dis(const Point &p1, const Point &p2) {
	return sqrt(1.0*(p1.x - p2.x)*(p1.x - p2.x) + 1.0*(p1.y - p2.y)*(p1.y - p2.y));
}

int fa[maxn];
int find(int x) { return fa[x] = fa[x] == x ? x : find(fa[x]); }

void init() {
	for (int i = 0; i <= n; i++) fa[i] = i;
}

int main() {
	while (scanf("%d", &n) == 1 && n) {
		init();
		int tot = 0;
		for (int i = 0; i < n; i++) {
			scanf("%d%d", &pt[i].x, &pt[i].y);
			for (int j = 0; j < i; j++) {
				egs[tot++] = Edge(j, i, dis(pt[j], pt[i]));
			}
		}
		sort(egs, egs + tot);
		double ans = 0;
		for (int i = 0; i < tot; i++) {
			Edge& e = egs[i];
			int pu = find(e.u);
			int pv = find(e.v);
			if (pu != pv) {
				ans += e.w;
				fa[pv] = pu;
			}
		}
		printf("%.2lf
", ans);
	}
	return 0;
}