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;

}