北京大学机试 Freckles 需要二刷 *最小生成树问题

基本思想：

本质上是利用一次稠密图Prim算法解决，不难；

 

关键点：

无；

 

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

const int maxn = 110;
const double INF = 1000000000.0;

double ma[maxn][maxn];
bool vis[maxn];
double dis[maxn];

struct node{
	double x, y;
};

vector<node>vec;

int n;

void init() {
	fill(dis, dis + maxn, INF);
	fill(ma[0], ma[0] + maxn * maxn, INF);
	fill(vis, vis + maxn, false);
}

double prim() {
	dis[0] = 0;
	double ans=0;
	for (int u = 0; u < n; u++) {
		int index = -1;
		double min = INF;
		for (int i = 0; i < n; i++) {
			if (!vis[i] && min > dis[i]) {
				min = dis[i];
				index = i;
			}
		}
		if (index == -1)
			return -1;
		ans += dis[index];
		vis[index] = true;
		for (int i = 0; i < n; i++) {
			if (!vis[i] && ma[index][i] != INF && ma[index][i] < dis[i]) {
				dis[i] = ma[index][i];
			}
		}
	}
	return ans;
}

int main() {
	while (cin >> n) {
		double a, b;
		init();
		for (int i = 0; i < n; i++) {
			cin >> a >> b;
			node no;
			no.x = a;
			no.y = b;
			vec.push_back(no);
			for (int i = 0; i < vec.size()-1; i++) {
				double dis = pow(a - vec[i].x,2) + pow(b - vec[i].y,2);
				dis = sqrt(dis);
				ma[i][vec.size() - 1] = ma[vec.size() - 1][i] = dis;
			}
		}
		printf("%.2lf\n", prim());
	}
	return 0;
}