PAT程序设计考题——甲级1003（Emergency ） C++实现

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#include<iostream>
#include<math.h>
#include<algorithm>
#include<queue>
#include<map>
#include<stack>
#include<string>
#include<vector>
using namespace std;
#define INF 100000000
#define maxn 100010
struct edge{
 int end,weight;
};
struct node{
 int pop;
 vector<edge> bian;
};
vector<node> ad;
int w[maxn];//记录点权之和
int pre[maxn];
int vt[maxn];
vector<int>  result;
int accord[maxn];//记录到到每个点最短路径的条数
bool visit[maxn]={false};
void Dji(int v,int num){
 fill(vt,vt+num,INF);
 fill(w,w+num,0);
 fill(accord,accord+num,0);
 vt[v]=0;
 accord[v]=1;
 w[v]=ad[v].pop;
// cout<<w[v];
 for(int i=0;i<num;i++)
 {
  int u=-1,store=INF;
  for(int j=0;j<num;j++)
  {
   if(vt[j]<store&&visit[j]==false){
    u=j;
    store=vt[j];
   }
  }
  if(u==-1) return;
  visit[u]=true;
  for(int k=0;k<ad[u].bian.size();k++)
  { int z;
  z=ad[u].bian[k].end;
  if(visit[z]==false){
  if(vt[u]+ad[u].bian[k].weight<vt[z])
  {
  vt[z]=vt[u]+ad[u].bian[k].weight;
  accord[z]=accord[u];
  w[z]=w[u]+ad[z].pop;}
  else if(vt[u]+ad[u].bian[k].weight==vt[z])
  { if(w[u]+ad[z].pop>w[z])
  w[z]=w[u]+ad[z].pop;
  accord[z]+=accord[u]; //注意路径的写法 
  }
  }
  }
  } 
}
int main()
{int city,c1,c2,num;
cin>>city>>num>>c1>>c2;
int pop;
for(int i=0;i<city;i++)
{
 node temp;
 cin>>pop;
 temp.pop = pop;
 ad.push_back(temp);
}
for (int i = 0; i < num; i++)
{
 int start, end, weight;
 cin >> start >> end >> weight;
 edge qw;
 qw.end = end; qw.weight = weight;
 ad[start].bian.push_back(qw);
 qw.end=start;
 ad[end].bian.push_back(qw);
}
Dji(c1,city);
/*for(int i=0;i<ad[4].bian.size();i++)
cout<<ad[4].bian[i].end<<" "<<ad[4].bian[i].weight<<endl;*/
cout<<accord[c2]<<" "<<w[c2];
 return 0;
}