Article From:

Title Link:

Problem Description
In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.
The transport system is very special: all
lines are unidirectional and connect exactly two stops. Buses leave the
originating stop with passangers each half an hour. After reaching the
destination stop they return empty to the originating stop, where they
wait until the next full half an hour, e.g. X:00 or X:30, where ‘X’
denotes the hour. The fee for transport between two stops is given by
special tables and is payable on the spot. The lines are planned in such
a way, that each round trip (i.e. a journey starting and finishing at
the same stop) passes through a Central Checkpoint Stop (CCS) where each
passenger has to pass a thorough check including body scan.

the ACM student members leave the CCS each morning. Each volunteer is
to move to one predetermined stop to invite passengers. There are as
many volunteers as stops. At the end of the day, all students travel
back to CCS. You are to write a computer program that helps ACM to
minimize the amount of money to pay every day for the transport of their



input consists of N cases. The first line of the input contains only
positive integer N. Then follow the cases. Each case begins with a line
containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is
the number of stops including CCS and Q the number of bus lines. Then
there are Q lines, each describing one bus line. Each of the lines
contains exactly three numbers – the originating stop, the destination
stop and the price. The CCS is designated by number 1. Prices are
positive integers the sum of which is smaller than 1000000000. You can
also assume it is always possible to get from any stop to any other


each case, print one line containing the minimum amount of money to be
paid each day by ACM for the travel costs of its volunteers.


Sample Input
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50


Sample Output
  1 /*
  2 problem  3 The number of the input vertices P and the edge number Q and the Q edge  4 Calculate and output the shortest path cost from vertex 1 to each vertex, plus the shortest path cost per vertex to 1.  5 
  6 The idea of solving the problem  7 It is not difficult to find the shortest path of the 1 vertex to the other vertex, and the shortest path of the single source can be solved by Dijkstra, and the key is the sum of the shortest path of the other vertices to the 1 vertex.  8 All sides of the original map can be stored in reverse, and the Dijkstra single source shortest path of the 1 vertex to the other vertex is the other vertex to the top 1.  9 The sum of the shortest path of the point. 10 In addition, because of the many vertices and edges, the Dijkstra algorithm based on adjacency list + priority queue optimization is adopted. 11 */ 
 12 #include<bits/stdc++.h>//HDU G++ 
 13 const int maxn=1e6+7;
 14 const int INF=1e9+7;
 16 using namespace std;
 18 int u[maxn],v[maxn],w[maxn]; 
 20 struct Edge{
 21     int from,to,dist;
 22 }; 
 24 struct HeapNode{
 25     int d,u;
 26     bool operator < (const HeapNode& rhs) const {//Priority queue, overloaded < operator
 27         return d >rhs.d;
 28     }
 29 };
 31 struct Dijkstra{
 32     int n,m;
 33     vector<Edge> edges;  //Adjacency list
 34     vector<int> G[maxn]; //The number of nodes starting from each node (numbered from 0)
 35     bool done[maxn];     //Whether or not it has been numbered permanently
 36     int d[maxn];         //sDistance to each point
 37     int p[maxn];         //The upper edge of the shortest path
 39     void init(int n){
 40         this->n =n;
 41         for(int i=0;i<n;i++)    G[i].clear();//Empty adjacency list
 42         edges.clear();                         //Emptying edge table
 43     }
 45     void AddEdge(int from,int to,int dist){
 46         //If it is an undirected graph, each side must be stored on both sides, and the two time AddEdge is called.
 47         edges.push_back((Edge){from,to,dist});
 48         m=edges.size();
 49         G[from].push_back(m-1);
 50     }
 52     void dijkstra(int s){//The distance from s to other points
 53         priority_queue<HeapNode> Q;
 54         for(int i=0;i<n;i++)    d[i]=INF; 
 55         d[s]=0;
 57         memset(done,0,sizeof(done));
 58         Q.push((HeapNode){0,s});
 60         while(!Q.empty()){
 61             HeapNode x;
 62             Q.pop();
 64             int u=x.u;
 65             if(done[u]) continue;
 66             done[u]=true;
 68             for(int i=0;i<G[u].size();i++){
 69                 Edge& e = edges[G[u][i]];
 70                 if(d[] > d[u] + e.dist){
 71                     d[] = d[u] + e.dist;
 72                     p[] = G[u][i];
 73                     Q.push((HeapNode){d[],});
 74                 }
 75             }
 76         }
 77     }
 78 };
 80 struct Dijkstra solver;
 82 int main() 
 83 {
 84     int T,n,m;
 85     scanf("%d",&T);
 86     while(T--){
 87         scanf("%d%d",&n,&m);
 88         solver.init(n);
 89         for(int i=1;i<=m;i++){
 90             scanf("%d%d%d",&u[i],&v[i],&w[i]);
 91             u[i]--;//The vertices in the template start from 0
 92             v[i]--;
 93             solver.AddEdge(u[i],v[i],w[i]);
 94         }
 95         solver.dijkstra(0);
 96         int ans=0;
 97         for(int i=0;i<solver.n;i++)
 98             ans += solver.d[i];
100         solver.init(n);
101         for(int i=1;i<=m;i++){
102             solver.AddEdge(v[i],u[i],w[i]);//Reverse storage after emptying
103         } 
104         solver.dijkstra(0);
105         for(int i=0;i<solver.n;i++)
106             ans += solver.d[i];
108         printf("%d\n",ans);
109     }
110     return 0;
111 }



Leave a Reply

Your email address will not be published. Required fields are marked *