hdu1535 Invitation Cards 最短路

有一张图，若干人要从不同的点到同一个中间点，再返回，求总费用最小

中间点到各个点最小费用是普通的最短路

各个点到中间点最小费用其实就是将所有路径反向建边之后中间点到各个点的最小费用，同样用最短路就可以求出了

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef pair<int,int> pii;
 typedef long long ll;
 const int MAXM=;
 struct cmp{
     bool operator ()(pii a,pii b){
         return a.first>b.first;
     }
 };

 int head1[MAXM+],point1[MAXM+],val1[MAXM+],next1[MAXM+],size1;
 int head2[MAXM+],point2[MAXM+],val2[MAXM+],next2[MAXM+],size2;
 int n,m;
 int dist1[MAXM+],dist2[MAXM+];

 void add1(int a,int b,int v){
     point1[size1]=b;
     val1[size1]=v;
     next1[size1]=head1[a];
     head1[a]=size1++;
 }

 void add2(int a,int b,int v){
     point2[size2]=b;
     val2[size2]=v;
     next2[size2]=head2[a];
     head2[a]=size2++;
 }

 void dij(int s){
     int i;
     priority_queue<pii,vector<pii>,cmp>q;
     memset(dist1,0x3f,sizeof(dist1));
     memset(dist2,0x3f,sizeof(dist2));
     dist1[s]=dist2[s]=;
     q.push(make_pair(dist1[s],s));
     while(!q.empty()){
         pii u=q.top();
         q.pop();
         if(u.first>dist1[u.second])continue;
         for(i=head1[u.second];~i;i=next1[i]){
             int j=point1[i];
             if(dist1[j]>dist1[u.second]+val1[i]){
                 dist1[j]=dist1[u.second]+val1[i];
                 q.push(make_pair(dist1[j],j));
             }
         }
     }
     while(!q.empty())q.pop();
     q.push(make_pair(dist2[s],s));
     while(!q.empty()){
         pii u=q.top();
         q.pop();
         if(u.first>dist2[u.second])continue;
         for(i=head2[u.second];~i;i=next2[i]){
             int j=point2[i];
             if(dist2[j]>dist2[u.second]+val2[i]){
                 dist2[j]=dist2[u.second]+val2[i];
                 q.push(make_pair(dist2[j],j));
             }
         }
     }
     ll ans=;
     for(i=;i<=n;i++){
         ans+=dist1[i];
         ans+=dist2[i];
     }
     printf("%I64d\n",ans);
 }

 int main(){
     int t;
     scanf("%d",&t);
         for(int q=;q<=t;q++){
             scanf("%d%d",&n,&m);
             int i,a,b,v;
             memset(head1,-,sizeof(head1));
             size1=;
             memset(head2,-,sizeof(head2));
             size2=;
             for(i=;i<=m;i++){
                 scanf("%d%d%d",&a,&b,&v);
                 add1(a,b,v);
                 add2(b,a,v);
             }
             dij();
         }

     return ;
 }