今天在牛客網上過國慶節的時候發現一個有點意思的最短路問題，連結在上面，程式碼在下面：

#include<bits/stdc++.h>
using namespace std;
int a,b,c1,c2;
const int INF =1e8; 
struct node{
	int x,y,r;
}cir[1005];
int book[1005];
double distance1(node A,node B){
	double dis=sqrt(pow(A.x-B.x,2)+pow(A.y-B.y,2));
	return max(dis-A.r-B.r,0.0);
}
double distance2(node A,int c){
	double dis=abs(a*A.x+b*A.y+c)/sqrt(pow(a,2)+pow(b,2));
	return max(dis-A.r,0.0);
}
double dis[1005];
int main()
{
	int n;
	cin>>n>>a>>b>>c1>>c2;
	memset(dis,1e5,sizeof(dis));
	memset(book,0,sizeof(book));
	for(int i=0;i<n;i++){
		scanf("%d%d%d",&cir[i].x,&cir[i].y,&cir[i].r);
	}
	for(int i=0;i<n;i++){
		dis[i]=distance2(cir[i],c1);
	}
	dis[n]=abs(c1-c2)/sqrt(pow(a,2)+pow(b,2));
	int u;
	for (int i=1;i<=n;i++){
		int min=INF;
	 	for (int j=0;j<n;j++){
			if (book[j]==0 && dis[j]<min){			//找出未確定最短路徑的距離源點最近的點 
				min=dis[j];
				u=j;
			}
		}
		book[u]=1;	//標記u點    說明到u點的最短路徑已經確定 
		for (int v=0;v<n;v++){
			if (dis[v]>dis[u]+distance1(cir[u],cir[v]))
			dis[v]=dis[u]+distance1(cir[u],cir[v]);
		}
		if(dis[n]>dis[u]+distance2(cir[u],c2))
			dis[n]=dis[u]+distance2(cir[u],c2);
	}	 
	printf("%.6lf",dis[n]);
}