Problem Description

There are m people in n cities, and they all want to attend the party which hold by zz. They set out at the same time, and they all will choose the best way they think, but due to someone take a ride, someone drive, and someone take a taxi, they have different speed. Can you find out who will get zz's mysterious present? The first one get the party will get the present . If there are several people get at the same time, the one who stay in the city which is farther from the city where is zz at begin will get the present. If there are several people get at the same time and the distance from the city he is at begin to the city where zz is, the one who has the larger number will get the present.

Input

The first line: three integers n, m and k. m is the total number of the people, and n is the total number of cities, and k is the number of the way.(0<n<=300, 0<m<=n, 0<k<5000)
The second line to the (k+1)th line: three integers a, b and c. There is a way from a to b, and the length of the way is c.(0<a,b<=n, 0<c<=100)
The (k+2)th line: one integer p(0<p<=n), p is the city where zz is.
The (k+3)th line: m integers. the ith people is at the place p[i] at begin.(0<p[i]<=n)
The (k+4)th line: m integers. the speed of the ith people is speed[i];(0<speed[i]<=100) 
All the ways are directed.

Output

For each case, output the one who get the present in one line. If no one can get the present, output "No one".

Sample Input

3 1 3
1 2 2
1 3 3
2 3 1
3
2
1

Sample Output

1

————————————————————————————————————————————————————

題意：給出一個 n 個點的有向圖，m 個人要到達點 p，每個人都有各自的速度，求最先到達 p 點的人的編號，如果多人同時到達，優先輸出初始點離 p 最遠的，如果多個人初始距離同樣遠，優先輸出編號最大的。

思路：建一個反向圖，求點 p 到個人的距離，從而可以得出每個人到達的時間，最後遍歷所有人的資訊，按花費時間最少、初始距離最遠、初始編號最大進行排序，最後輸出即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
using namespace std;
struct Node{
    int speed;
    int place;
    int id;
    int dis;
    double time;
}a[N];
int n,m,k;
int G[N][N];
int dis[N];
bool vis[N];
void Dijkstra(int s){
    memset(vis,false,sizeof(vis));
    vis[s]=true;
    for(int i=1;i<=n;i++)
        if(i!=s)
            dis[i]=G[s][i];
    dis[s]=0;

    for(int i=1;i<=n-1;i++){
        int minn=INF;
        int x=-1;
        for(int j=1;j<=m;j++){
            if(vis[j])
                continue;
            if(minn>dis[j]){
                minn=dis[j];
                x=j;
            }
        }

        if(x==-1)
            break;
        vis[x]=true;
        for(int j=1;j<=n;j++)
            if(!vis&&dis[j]>dis[x]+G[x][j])
                dis[j]=dis[x]+G[x][j];
    }
}
bool cmp(Node a,Node b){
    if(fabs(a.time-b.time)>E)
        return a.time>b.time;
    else if(a.dis!=b.dis)
        return a.dis>b.dis;
    else
        return a.id>b.id;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF&&(n+m+k))
    {
        memset(G,INF,sizeof(G));
        for(int i=1;i<=n;i++)
            G[i][i]=0;

        for(int i=1;i<=k;i++){
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            G[y][x]=min(G[y][x],w);
        }

        int start;
        scanf("%d",&start);
        Dijkstra(start);
        for(int i=1;i<=m;i++){
            scanf("%d",&a[i].place);
            a[i].id=i;
            a[i].dis=dis[a[i].place];
        }

        for(int i=1;i<=m;i++){
            scanf("%d",&a[i].speed);
            if(dis[a[i].place]==INF)
                a[i].time=INF;
            else
                a[i].time=1.0*a[i].dis/a[i].speed;
        }

        sort(a+1,a+1+m,cmp);
        if(a[1].dis==INF)
            printf("No one\n");
		else
            printf("%d\n",a
                   [1].id);

    }
    return 0;
}