poj1511 最短路

题意：与poj3268一样，所有人需要从各点到一点再从一点到各点，求最短路总和。

与poj3268一样，先正向建图跑 dijkstra ，得到该点到其他所有各点的最短路，即这些人回去的最短路，再用反向建的图跑一遍最短路，得到各点到该点的最短路，求和就行了。

 #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 ;
 }