A gas station has to be built at such a location that the minimum distance between the station and any of the residential housing is as far away as possible. However it must guarantee that all the houses are in its service range.

Now given the map of the city and several candidate locations for the gas station, you are supposed to give the best recommendation. If there are more than one solution, output the one with the smallest average distance to all the houses. If such a solution is still not unique, output the one with the smallest index number.

Input Specification:
Each input file contains one test case. For each case, the first line contains 4 positive integers: N (≤10
​3
​​ ), the total number of houses; M (≤10), the total number of the candidate locations for the gas stations; K (≤10
​4
​​ ), the number of roads connecting the houses and the gas stations; and D
​S
​​ , the maximum service range of the gas station. It is hence assumed that all the houses are numbered from 1 to N, and all the candidate locations are numbered from G1 to GM.

Then K lines follow, each describes a road in the format

P1 P2 Dist
where P1 and P2 are the two ends of a road which can be either house numbers or gas station numbers, and Dist is the integer length of the road.

Output Specification:
For each test case, print in the first line the index number of the best location. In the next line, print the minimum and the average distances between the solution and all the houses. The numbers in a line must be separated by a space and be accurate up to 1 decimal place. If the solution does not exist, simply output No Solution.

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Sample Input 1:
4 3 11 5
1 2 2
1 4 2
1 G1 4
1 G2 3
2 3 2
2 G2 1
3 4 2
3 G3 2
4 G1 3
G2 G1 1
G3 G2 2
Sample Output 1:
G1
2.0 3.3
Sample Input 2:
2 1 2 10
1 G1 9
2 G1 20
Sample Output 2:
No Solution

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#include<iostream>
#include<vector>
#include<cstring>
#include<cmath>
#define INF 0x3f3f3f
#define MAX 1020
using namespace std;
int n,m,k,ds;
int dis[MAX],g[MAX][MAX];
vector<int> pre[MAX],tmp,path;
int vis[MAX];
int mini=INF,lremain=INF;
#define INF 0x3f3f3f
void dij(int src){
	fill(dis,dis+MAX,INF);
	memset(vis,0,sizeof(vis));
	dis[src]=0;
	//循环n+m次 每一次占一个点（城堡） V S-v集合 
	for(int i=0;i<n+m;i++){
		int min=INF,u=-1;
		for(int j=1;j<=n+m;j++){
			if(!vis[j]&&dis[j]<min){
				min=dis[j];u=j;
			}
		}
		if(u==-1) return;
		vis[u]=true;
		for(int v=1;v<=n+m;v++){
			if(vis[v]||g[u][v]==INF) continue;
			if(dis[u]+g[u][v]<dis[v]){
				dis[v]=dis[u]+g[u][v];
				pre[v].clear();
				pre[v].push_back(u);
			}
			else if(dis[u]+g[u][v]==dis[v]){
					pre[v].push_back(u);
			}
		}
		
		
	}
	
}
void dfs(int u){
}
int getId(string s){
	int ans=0;
	bool flag=false;
	for(int i=0;i<s.length();i++){
		if(s[i]=='G'){
			flag=true;
		}
		else ans=ans*10+s[i]-'0';
	}
	if(flag==true) return n+ans;
	else return ans;

} 
int main(){
//	freopen("21.txt","r",stdin);
	fill(g[0],g[0]+MAX*MAX,INF);
	cin>>n>>m>>k>>ds; 
	int ansid=-1;
	double ansA=INF,ansL=-1;
	for(int i=0;i<k;i++)
	{
		string s1,s2;
		int d;
		cin>>s1>>s2>>d;
		int u=getId(s1),v=getId(s2);
		g[u][v]=g[v][u]=d;
	}
	for(int i=n+1;i<=n+m;i++){
		dij(i);
		int  min=INF,u=-1;
		double avg=0;
		for(int j=1;j<=n;j++){
			if(dis[j]<min){
				u=j;
				min=dis[j];
			}
			if(dis[j]>ds) {
				min=-1;
				break;
			}
			avg+=1.0*dis[j]/n;
		}
//		cout<<min<<" "<<avg<<endl;
		if(min==-1) continue;
//		if(min>ds) {
//			continue;
//		}
		

		if(min>ansL){
			ansid=i; 
			ansL=min;
			ansA=avg;
		} 
		else if(min==ansL){
			if(avg<ansA){
				ansid=i; //!
				ansA=avg;
			}
		}
		
		
			
	}
	
	if(ansid==-1) cout<<"No Solution"<<endl;
	else{
		cout<<'G'<<ansid-n<<endl;
		printf("%.1f %.1f\n",ansL,ansA);
	}
}

There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any station and return it to any other stations in the city.

A long-distance telephone company charges its customers by the following rules:

A traveler’s map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all his/her followers can view and forward his/her post, which can then be forwarded again by their followers. Now given a social network, you are supposed to calculate the maximum potential amount of forwards for any specific user, assuming that only L levels of indirect followers are counted.

“答案正确”是自动判题系统给出的最令人欢喜的回复。本题属于 PAT 的“答案正确”大派送 —— 只要读入的字符串满足下列条件，系统就输出“答案正确”，否则输出“答案错误”。

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

One way that the police finds the head of a gang is to check people’s phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A “Gang” is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threthold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.