洛谷P1027 Car的旅行路线

版权声明：我这么弱的蒟蒻，虽然博文不是很好，但也请标明转发地址喵！ https://blog.csdn.net/ACerAndAKer/article/details/81866348

几何+最短路

我们根据已给的三个坐标，通过勾股定理，确认第四个点，然后SPFA时判断是不是同一个城市就行了

代码

//By AcerMo
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int M=1050;
int w[M];
bool vis[M];
int n,m,st,ed;
double dis[M];
int x[M],y[M],id[M],cnt;
inline int read()
{
	int x=0;char ch=getchar();
	while (ch>'9'||ch<'0') ch=getchar();
	while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
inline int calc(int x,int y)
{
	return (x-y)*(x-y);
}
inline void get4(int i)
{
	int a=calc(x[cnt],x[cnt-1])+calc(y[cnt],y[cnt-1]);
	int b=calc(x[cnt],x[cnt-2])+calc(y[cnt],y[cnt-2]);
	int c=calc(x[cnt-1],x[cnt-2])+calc(y[cnt-1],y[cnt-2]);
	int x4,y4;
	if (a+b==c) x4=x[cnt-1]+x[cnt-2]-x[cnt],y4=y[cnt-1]+y[cnt-2]-y[cnt];
	if (a+c==b) x4=x[cnt]+x[cnt-2]-x[cnt-1],y4=y[cnt]+y[cnt-2]-y[cnt-1];
	if (b+c==a) x4=x[cnt]+x[cnt-1]-x[cnt-2],y4=y[cnt]+y[cnt-1]-y[cnt-2];
	x[++cnt]=x4;y[cnt]=y4;id[cnt]=i;
	return ;
}
inline void SPFA()
{
	fill(dis,dis+M,2e9);
	queue<int>q;dis[st*4]=dis[st*4-1]=dis[st*4-2]=dis[st*4-3]=0;
	q.push(st*4),q.push(st*4-1),q.push(st*4-2),q.push(st*4-3);
	while (q.size())
	{
		int u=q.front();q.pop();vis[u]=0;
		for (int i=1;i<=4*n;i++)
		{
			if (i==u) continue;
			double cost=(double)sqrt(calc(x[u],x[i])+calc(y[u],y[i]));
			if (id[u]==id[i]) cost*=(double)w[id[i]];
			else cost*=(double)m;
			if (dis[i]>dis[u]+cost)
			{
				dis[i]=dis[u]+cost;
				if (!vis[i]) 
					vis[i]=1,q.push(i);
			}
		}
	}
	return ;
}
signed main()
{
	int t=read();
	while (t--)
	{
		n=read();m=read();st=read();ed=read();
		for (int i=1;i<=n;i++)
		{
			x[++cnt]=read(),y[cnt]=read();id[cnt]=i;
			x[++cnt]=read(),y[cnt]=read();id[cnt]=i;
			x[++cnt]=read(),y[cnt]=read();id[cnt]=i;
			get4(i);w[i]=read();id[cnt]=i;
		}
		SPFA();double ans=2e9;
		for (int i=0;i<=3;i++)
		ans=min(ans,dis[ed*4-i]);
		printf("%.1lf\n",ans);
	}
	return 0;
}