HDU 携程编程大赛 1003 携程全球数据中心建设 Prim OR Kruskal-优快云博客

本文链接：https://blog.youkuaiyun.com/rockingbee/article/details/23386157

本文介绍了一种利用最小生成树算法解决特定问题的方法。通过计算二维平面上各点之间的距离并构建最小生成树来判断是否满足给定条件。文章详细展示了使用C++实现的全过程，包括距离计算和Prim算法的具体实现。

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水题

先求两两间的长度，再求最小生成树

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define BUG puts("HERE!")
using namespace std;
const int MAX = 105;
const double INF = 0x7fffff;
const double eps = 1e-5;
const double PI = 3.14159265358979323846;
double dis[MAX][MAX], x[MAX], y[MAX];
double dist[MAX];
int t, c;
double d, l;
bool vis[MAX];

double prim(int n) {  
	int i,j,pos;  
	double sum=0,min;
	memset(vis,false,sizeof(vis));
	for(i=1; i<=n; i++) {  
		dist[i]=dis[1][i];  
	}
	vis[1] = true; 
	dist[1] = INF;
	for(i=1; i<n; i++) {  
		min = INF;
		for(j = 1; j <= n; j++) {  
			if(!vis[j] && dist[j] < min) {  
				min = dist[j];  
				pos = j;  
			}  
		}  
		sum += dist[pos];
		vis[pos] = true; 
		for(j = 1; j <= n; j++) {  
			if(!vis[j] && dist[j] > dis[pos][j]) {  
				dist[j] = dis[pos][j];
			}  
		}  
	}  
	return sum;  
} 

double dd(double x1, double y1, double x2, double y2){
	return fabs(d / 2  *acos(sin(x1)*sin(x2)+cos(x1)*cos(x2)*cos(y1-y2)));
}

int main(int argc, char const *argv[]){

#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
	// freopen("out.out", "w", stdout);
#endif

	cin >> t;
	while (t --){
		cin >> d >> l >> c;
		for (int i = 1; i <= c; ++i){
			cin >> x[i] >> y[i];
			x[i] = x[i] * PI / 180.0;
			y[i] = y[i] * PI / 180.0;
		}
		for (int i = 1; i <= c; ++i){
			for (int j = 1; j <= c; ++j){
				if(i == j)
					dis[i][j] = 0;
				else
					dis[i][j] = dd(x[i], y[i], x[j], y[j]);
			}
		}
		double ans = prim(c);
		if(ans < l)
			puts("Y");
		else
			puts("N");
	}
	return 0;
}