#include<string.h>
#include<fstream>
#include <iostream>
#include<queue>
using namespace std;
const int MAXN=500;
const int INF=0xfffff;

struct City
{
public:

int cityNumber;
int cost;
City*next;
};
struct Atr
{
public:
int cityIndex;
int teamCount;
City* link;
};
class SaveCity
{
public:
void dijstra(Atr*atr,int n,int src,int des);
};

//每个顶点的父亲节点，可以用于还原最短路径树
bool*visited; //用于判断顶点是否已经在最短路径树中，或者说是否已找到最短路径
int*dist;
//存储的是到该节点的路径数
int* roadCount;
//存储的是最多救援队的路线
int* maxTeamSum;
int*parent;
struct cmp
{
bool operator ()(int &a,int &b){
return dist[a]>dist[b];//最小值优先
}
};
priority_queue<int,vector<int>,cmp> q; //优先队列stl实现

int main()
{
//构造图的边链表存储数据结构
Atr* atr;
int n,m,src,des;

// ifstream infile;
// infile.open("data.txt",ios::in);
cin>>n>>m>>src>>des;
// infile>>n>>m>>src>>des;
atr=new Atr[n];
for(int i=0;i<n;i++)
{
atr[i].cityIndex=i;
cin>>atr[i].teamCount;
//infile>>atr[i].teamCount;
atr[i].link=NULL;
}
for(int i=0;i<m;i++)
{

int temp=-1;
cin>>temp;
//infile>>temp;
City*city=new City();

cin>>city->cityNumber;
//infile>>city->cityNumber;
cin>>city->cost;
//infile>>city->cost;
city->next=NULL;

if(atr[temp].link==NULL)
{
atr[temp].cityIndex=temp;

atr[temp].link=city;
}
else
{
City* city1= atr[temp].link;
while(city1->next!=NULL)
{
city1=city1->next;
}
city1->next=city;
}
}

//找最短路径
SaveCity* saveCity=new SaveCity;
saveCity->dijstra(atr,n,src,des);
delete[] atr;
delete saveCity;
// infile.close();
return 0;
}
void SaveCity::dijstra(Atr*atr,int n,int src,int des)
{
//动态开辟空间
parent=new int[n];
visited=new bool[n];
dist=new int[n];
roadCount=new int[n];
maxTeamSum=new int[n];
//初始化
for(int i=0;i<n;i++)
{
parent[i]=-1;//每个顶点都没有父节点
visited[i]=false;//每个节点都没有访问过
dist[i]=INF;//每个距离都是很大的值
roadCount[i]=1;
maxTeamSum[i]=0;
}
maxTeamSum[src]=atr[src].teamCount;
dist[src]=0;
q.push(src);
while(!q.empty())
{
//弹出一个节点,优先队列每次都弹出最小dist的节点
int top=q.top();
//cout<<"*******************"<<top<<endl;
q.pop();
if(visited[top])
{
continue;
}
//松弛操作//找所有与他相邻的顶点，进行松弛操作，
//更新估算距离，压入队列。
City* city=atr[top].link;
while(city!=NULL)
{
int cityNum=city->cityNumber;
if((!visited[cityNum] )&& dist[cityNum] >=dist[top]+city->cost)
{
if(dist[cityNum] >dist[top]+city->cost)
{
dist[cityNum] = dist[top]+city->cost;

parent[cityNum] = top;
maxTeamSum[cityNum]=maxTeamSum[top]+atr[cityNum].teamCount;
q.push(cityNum);
}
else
{
if(maxTeamSum[top]>maxTeamSum[parent[cityNum]])
{
parent[cityNum] = top;
maxTeamSum[cityNum]=maxTeamSum[top]+atr[cityNum].teamCount;
}
parent[cityNum] = top;
roadCount[cityNum]++;
}
}
city = city->next;
}

visited[top]=true;
}
//输出结果SaveCity
cout<<roadCount[des]<<" "<<maxTeamSum[des];
return;

}

Pat-1003