poj2253Frogger

/*poj2253Frogger
题意：青蛙要通过到石头到达终点 要达到终点最小步幅不能小于多少
dijkstra
*/

#include<iostream>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define MAXN 1111
#define INF 100000000
pair <double,int> pii;
double x[MAXN],y[MAXN],d[MAXN];
int inq[MAXN];
int n;
double DIS(int i,int j)
{
	return sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
void dijkstra()
{
	//priority_queue<pii>pq;
	queue<int>q;
    memset(inq,0,sizeof(inq));
	for(int i=0;i<n;i++)d[i]=(i==0?-INF:INF);
	//pq.push(make_pair(INF,))
	q.push(0);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		inq[u]=0;
		for(int i=0;i<n;i++)
		{
			double len=max(DIS(u,i),d[u]);//更改判断条件
			if(len<d[i])
			{
				d[i]=len;
				if(!inq[i])
				{
					inq[i]=1;
					q.push(i);
				}
			}
		}
	}
}
int main()
{
	cout.precision(3);
	int cas=0;
	while(scanf("%d",&n)==1)
	{
		if(!n)break;
		for(int i=0;i<n;i++)
		{
			scanf("%lf%lf",&x[i],&y[i]); 
		}
		dijkstra();
		printf("Scenario #%d\nFrog Distance = ",++cas);
		cout<<fixed<<d[1]<<endl;
		printf("\n");
	}
}