題目連接：php
http://acm.hdu.edu.cn/showproblem.php?pid=1535c++
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.
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 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 employees.
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.
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 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 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
2
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
46
210
1 /* 2 問題 3 輸入頂點數p和邊數q以及q條邊 4 計算並輸出頂點1到每一個頂點的最短路徑花費，再加上每一個頂點到1的最短路徑花費 5 6 解題思路 7 求1號頂點到其他頂點的最短路徑之和不難，Dijkstra求解單源最短路便可，關鍵是其他頂點到1號頂點的最短路徑之和 8 具體作法是將原圖的全部邊都反向存儲一遍，再跑一邊1號頂點到其他頂點的Dijkstra單源最短路就是要求的其他頂點到1號頂 9 點的最短路徑之和。 10 另外因爲頂點和邊不少，因此採用鄰接表+優先隊列優化的Dijkstra算法。 11 */ 12 #include<bits/stdc++.h>//HDU G++ 13 const int maxn=1e6+7; 14 const int INF=1e9+7; 15 16 using namespace std; 17 18 int u[maxn],v[maxn],w[maxn]; 19 20 struct Edge{ 21 int from,to,dist; 22 }; 23 24 struct HeapNode{ 25 int d,u; 26 bool operator < (const HeapNode& rhs) const {//優先隊列，重載<運算符 27 return d >rhs.d; 28 } 29 }; 30 31 struct Dijkstra{ 32 int n,m; 33 vector<Edge> edges; //鄰接表 34 vector<int> G[maxn]; //每一個節點出發的邊編號（從0開始編號） 35 bool done[maxn]; //是否已經永久編號 36 int d[maxn]; //s到各個點的距離 37 int p[maxn]; //最短路中的上一條邊 38 39 void init(int n){ 40 this->n =n; 41 for(int i=0;i<n;i++) G[i].clear();//清空鄰接表 42 edges.clear(); //清空邊表 43 } 44 45 void AddEdge(int from,int to,int dist){ 46 //若是是無向圖須要將每條無向邊存儲兩邊，及調用兩次AddEdge 47 edges.push_back((Edge){from,to,dist}); 48 m=edges.size(); 49 G[from].push_back(m-1); 50 } 51 52 void dijkstra(int s){//求s到其餘點的距離 53 priority_queue<HeapNode> Q; 54 for(int i=0;i<n;i++) d[i]=INF; 55 d[s]=0; 56 57 memset(done,0,sizeof(done)); 58 Q.push((HeapNode){0,s}); 59 60 while(!Q.empty()){ 61 HeapNode x =Q.top(); 62 Q.pop(); 63 64 int u=x.u; 65 if(done[u]) continue; 66 done[u]=true; 67 68 for(int i=0;i<G[u].size();i++){ 69 Edge& e = edges[G[u][i]]; 70 if(d[e.to] > d[u] + e.dist){ 71 d[e.to] = d[u] + e.dist; 72 p[e.to] = G[u][i]; 73 Q.push((HeapNode){d[e.to],e.to}); 74 } 75 } 76 } 77 } 78 }; 79 80 struct Dijkstra solver; 81 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]--;//模板中的頂點從0開始 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]; 99 100 solver.init(n); 101 for(int i=1;i<=m;i++){ 102 solver.AddEdge(v[i],u[i],w[i]);//清空後反向存儲 103 } 104 solver.dijkstra(0); 105 for(int i=0;i<solver.n;i++) 106 ans += solver.d[i]; 107 108 printf("%d\n",ans); 109 } 110 return 0; 111 }