题意:球面上给出n个点的经纬度,求最小生成树总路径长度是否小于给定的一个长度;
解法:球面上的最小生成树,关键是有两点的经纬度得到两点的球面距离:
球面距离公式:length=R*acos(cosβ1*cosβ2*cos(α1-α2)+sinβ1*sinβ2),β1,β2分别为纬度,α1,α2分别为经度;
代码:
/****************************************************
* author:xiefubao
*******************************************************/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string.h>
using namespace std;
#define eps 1e-8
double pi=acos(-1);
typedef long long LL;
int parent[110];
int getparent(int k)
{
if(parent[k]==k)
return k;
return parent[k]=getparent(parent[k]);
}
struct p
{
double x,y;
} ps[110];
struct point
{
int a,b;
double length;
} points[100100];
int p=0;
bool operator<(point a,point b)
{
return a.length<b.length;
}
double R,L;
int C;
bool OK(int a,int b)
{
int ta=getparent(a);
int tb=getparent(b);
if(ta==tb)
return false;
parent[ta]=tb;
return true;
}
int main()
{
int t;cin>>t;
while(t--)
{
scanf("%lf%lf%d",&R,&L,&C);R/=2;
p=0;
for(int i=0;i<C;i++)
{
scanf("%lf%lf",&ps[i].x,&ps[i].y);
parent[i]=i;
}
for(int i=0;i<C;i++)
for(int j=i+1;j<C;j++)
{
double x1=ps[i].x/180*pi;
double x2=ps[j].x/180*pi;
double y1=ps[i].y/180*pi;
double y2=ps[j].y/180*pi;
points[p].a=i,points[p].b=j;
double tool=abs(y1-y2);
if(tool>pi)
tool-=pi;
points[p++].length=R*acos(cos(x1)*cos(x2)*cos(tool)+sin(x1)*sin(x2));
}
sort(points,points+p);
double sum=0;
for(int i=0;i<p;i++)
{
if(OK(points[i].a,points[i].b))
sum+=points[i].length;
}
if(sum<=L)
cout<<"Y\n";
else
cout<<"N\n";
}
return 0;
}
/*
R*arccos(cosx1*cosx2*cos(y1-y2)+sinx1*sinx2);
*/
携程编程赛第一场C题(球面最小生成树),布布扣,bubuko.com
原文:http://blog.csdn.net/xiefubao/article/details/23395823