POJ_1511_Invitation Cards (Shortest Path)

POJ_1511_Invitation Cards (Shortest Path)
Invitation Cards

Time Limit:8000 MS		Memory Limit:262144 K
Total Submissions:21615		Accepted:7089

Description

In the age of television, not every people attend theater has CES. 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 traveling 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 parameter 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.

All the ACM student members leave the CCS each morning. each volunteer is to move to one predetermined stop to invite passengers. there are as follows 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 employees.

Input

The 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 stop.

Output

For 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

22 21 2 132 1 334 61 2 102 1 601 3 203 4 102 4 54 1 50

Sample Output

 

46210

 

 

In a directed graph, calculate the sum of the shortest paths from 1 to all points + the sum of the shortest paths from all points to 1.

Analysis: it is obvious that the spfa or dijkstra + heap is used to find the shortest short path along the storage side, and then reverse the side to find the shortest path, and then sum the result. If this question uses vector to store graphs, it is easy to get stuck. For example, if I use two adjacent tables to store images directly, I changed it to a temporary table. However, it is still time-consuming and can be further optimized. The meaning of the question is clear. The number of edges is 1000000. Therefore, we can directly store the edges with struct arrays, so that operations on arrays will not take much time. For details, see the code:

Question link: http://poj.org/problem? Id = 1511

Code List:

Vector implementation (high time complexity ):

 

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# Define deusing namespace std; typedef long ll; const int maxn = 1000000 + 5; const int max_dis = 1e9 + 5; struct Edge {int to, dis; Edge (int, int dis) {this-> to = to; this-> dis = dis ;}}; struct edge {int from, to, dis ;}g [maxn]; int T; int n, Q; int a, B, c; int d [maxn]; bool vis [maxn]; typedef pair
      
        P; vector
       
         Graph [maxn]; void Clear () {for (int I = 1; I <= n; I ++) {graph [I]. clear () ;}} void input () {scanf (% d, & n, & Q); Clear (); for (int I = 0; I
        
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> Q; while (q. size () q. pop (); d [1] = 0; q. push (P (0, 1); while (q. size () {P p = q. top (); q. pop (); int v = p. second; if (d [v]
          
            D [v] + e. dis) {d [e. to] = d [v] + e. dis; q. push (P (d [e. to], e. to) ;}} ll ret = 0; for (int I = 1; I <= n; I ++) ret + = d [I]; return ret ;} ll spfa () {memset (vis, false, sizeof (vis); fill (d + 1, d + 1 + n, max_dis); queue
           
             Q; while (! Q. empty () q. pop (); d [1] = 0; vis [1] = true; q. push (1); while (! Q. empty () {int p = q. front (); q. pop (); vis [p] = 0; for (int I = 0; I
            
              D [p] + e. dis) {d [e. to] = d [p] + e. dis; if (! Vis [e. to]) {vis [e. to] = true; q. push (e. to) ;}}} ll ret = 0; for (int I = 1; I <= n; I ++) ret + = d [I]; return ret ;} void export stra_solve () {ll ans = dijkstra (); Clear (); for (int I = 0; I
             
               Struct Array Implementation (low time complexity ):
             
            
           
          
         
 
         
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               #includeusing namespace std;typedef long long ll;const int maxn=1000000+5;const int max_dis=1e9 + 5;struct Edge{int to,dis,next;}graph[maxn];struct edge{int from,to,dis;}g[maxn];int T;int n,Q;int num;int d[maxn];bool vis[maxn];int head[maxn];typedef pair
               
                P;void Clear(){ num=0; memset(head,-1,sizeof(head)); memset(graph,0,sizeof(graph));}void add(int u,int v,int dis){ graph[num].to=v; graph[num].dis=dis; graph[num].next=head[u]; head[u]=num++;}void input(){ scanf(%d%d,&n,&Q); Clear(); for(int i=0;i
                
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 >q; while(q.size()) q.pop(); d[1]=0; q.push(P(0,1)); while(q.size()){ P p=q.top(); q.pop(); int v=p.second; if(d[v]
                  
                   -1;k=graph[k].next){ if(d[graph[k].to]>d[v]+graph[k].dis){ d[graph[k].to]=d[v]+graph[k].dis; q.push(P(d[graph[k].to],graph[k].to)); } } } ll ret=0; for(int i=1;i<=n;i++) ret+=d[i]; return ret;}ll spfa(){ memset(vis,false,sizeof(vis)); fill(d+1,d+1+n,max_dis); queue
                   
                    q; while(!q.empty()) q.pop(); d[1]=0; vis[1]=true; q.push(1); while(!q.empty()){ int v=q.front(); q.pop(); vis[v]=0; for(int k=head[v];k>-1;k=graph[k].next){ if(d[graph[k].to]>d[v]+graph[k].dis){ d[graph[k].to]=d[v]+graph[k].dis; if(!vis[graph[k].to]){ vis[graph[k].to]=true; q.push(graph[k].to); } } } } ll ret=0; for(int i=1;i<=n;i++) ret+=d[i]; return ret;}void dijkstra_solve(){ ll ans=dijkstra(); Clear(); for(int i=0;i