Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33015		Accepted: 11174

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N 

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 51 2 202 3 303 4 204 5 201 5 100

Sample Output

90 理论上的效率应该是Dijsktra > SPFA > Bellman_Ford，但是前两者我用了vector，影响了效率，导致贝尔曼是最快的，迪杰斯特拉其次。

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 using    namespace    std; 6  7 const    int    SIZE = 1005; 8 const    int    INF = 0x2fffffff; 9 bool    S[SIZE];10 int    N,D[SIZE];11 struct    Node12 {13     int    vec,cost;14 };15 struct    comp16 {17     bool    operator ()(int & a,int & b)18     {19         return    D[a] > D[b];20     }21 };22 vector
          
               G[SIZE];23 priority_queue    
           
            
             ,comp> QUE;24 25 void dijkstra(int);26 void relax(int,int,int);27 int main(void)28 {29 int t,from;30 Node temp;31 32 scanf("%d%d",&t,&N);33 while(t --)34 {35 scanf("%d%d%d",&from,&temp.vec,&temp.cost);36 G[from].push_back(temp);37 swap(from,temp.vec);38 G[from].push_back(temp);39 }40 dijkstra(N);41 printf("%d\n",D[1]);42 43 return 0;44 }45 46 void dijkstra(int s)47 {48 fill(D,D + SIZE,INF);49 D[s] = 0;50 S[s] = true;51 QUE.push(s);52 53 while(!QUE.empty())54 {55 int cur = QUE.top();56 int len = G[cur].size();57 S[cur] = true;58 QUE.pop();59 for(int i = 0;i < len;i ++)60 relax(cur,G[cur][i].vec,G[cur][i].cost);61 if(cur == 1)62 return ;63 }64 }65 66 void relax(int from,int to,int cost)67 {68 if(D[to] > D[from] + cost)69 {70 D[to] = D[from] + cost;71 if(!S[to])72 QUE.push(to);73 }74 }
            
           
          
         
        
       
      

1 #include 
      
        2 #include 
       
         3 using    namespace    std; 4  5 const    int    INF = 0x5fffffff; 6 const    int    SIZE = 1005; 7 bool    UPDATE; 8 int    D[SIZE]; 9 int    N,E;10 struct    Node11 {12     int    from,to,cost;13 }Edge[SIZE * 4];14 15 void    Bellman_Ford(int);16 void    relax(int,int,int);17 int    main(void)18 {19     int    t;20     Node    temp;21 22     scanf("%d%d",&t,&N);23     while(t --)24     {25         scanf("%d%d%d",&temp.from,&temp.to,&temp.cost);26         Edge[E ++] = temp;27         swap(temp.from,temp.to);28         Edge[E ++] = temp;29     }30     Bellman_Ford(N);31     printf("%d\n",D[1]);32 33     return    0;34 }35 36 void    Bellman_Ford(int s)37 {38     fill(D,D + SIZE,INF);39     D[s] = 0;40 41     for(int i = 0;i < N - 1;i ++)42     {43         UPDATE = false;44         for(int j = 0;j < E;j ++)45             relax(Edge[j].from,Edge[j].to,Edge[j].cost);46         if(!UPDATE)47             return    ;48     }49 }50 51 void    relax(int from,int to,int cost)52 {53     if(D[to] > D[from] + cost)54     {55         D[to] = D[from] + cost;56         UPDATE = true;57     }58 }
       
      

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 using    namespace    std; 5  6 const    int    SIZE = 1005; 7 const    int    INF = 0x5fffffff; 8 int    N,D[SIZE]; 9 bool    IN_QUE[SIZE];10 struct    Node11 {12     int    to,cost;13 };14 vector
         
              G[SIZE];15 16 void    spfa(int);17 bool    relax(int,int,int);18 int    main(void)19 {20     int    t,from;21     Node    temp;22 23     scanf("%d%d",&t,&N);24     while(t --)25     {26         scanf("%d%d%d",&from,&temp.to,&temp.cost);27         G[from].push_back(temp);28         swap(from,temp.to);29         G[from].push_back(temp);30     }31     spfa(N);32     printf("%d\n",D[1]);33 34     return    0;35 }36 37 void    spfa(int s)38 {39     int    vec,cost;40     queue
          
               que;41     fill(D,D + SIZE,INF);42     D[s] = 0;43     IN_QUE[s] = true;44     que.push(s);45 46     while(!que.empty())47     {48         int    cur = que.front();49         int    len = G[cur].size();50         IN_QUE[cur] = false;51         que.pop();52 53         for(int i = 0;i < len;i ++)54         {55             vec = G[cur][i].to;56             cost = G[cur][i].cost;57             if(relax(cur,vec,cost) && !IN_QUE[vec])58             {59                 IN_QUE[vec] = true;60                 que.push(vec);61             }62         }63     }64 }65 66 bool    relax(int from,int to,int cost)67 {68     if(D[to] > D[from] + cost)69     {70         D[to] = D[from] + cost;71         return    true;72     }73     return    false;74 }