<poj-3268> Silver Cow Party Shortest Path problem

　　Link : http://poj.org/problem?id=3268

　　 The cows are going to the party, enter N = number of vertices (cattle field),M = number of sides (road), X = end point (Party point). Problem: The maximum value of the round trip time is asked.

Input:

Line 1:three space-separated integers, respectively: N, M, and X
Lines 2.. M+1:line I+1 describes road IWith three space-separated integers: Ai, Bi, and Ti. The described road runs from farm AiTo farm Bi, requiring TiTime units to traverse.

Output:

Line 1:one integer:the maximum of the time any one cow must walk.

　  4 8 2 1 2 4 1 3 2 1 4 7 2 1 1 2 3 5 3 1 2 3 4 4 4 2 3

Sample Output:

　　10

　　problem-solving ideas : Because the subject is one-way, so you can do a forward graph and a reverse graph, respectively, the time to solve the back and forth, and then find the maximum value.

Here is my code:

1#include <cstring>2#include <iostream>3 #defineINF 99999994 using namespacestd;5 6 BOOLused[100005];7 intV, E;8 Const intMAXN =1005;9 Ten voidDijkstra (intSintcost[][1005],intd[]) { OneFill (d, D + V +1, INF); AFill (used,used + V +1,false); -D[s] =0; -      while(true) { the         intv =-1; -          for(intU =1; U <= V; u++) { -             if(!used[u] && (v = =-1|| D[u] < d[v])) v =u; -         } +         if(v = =-1) Break; -USED[V] =true; +          for(intU =1; U <= V; ++u) { A             if(D[u] > D[v] +Cost[v][u]) { atD[u] = D[v] +Cost[v][u]; -             } -         } -     } - } -  in intCOST[MAXN][MAXN]; - intRCOST[MAXN][MAXN]; to  + intMain () { -     intD[MAXN]; the     intRD[MAXN]; *     intx, Y, W; $     intSUM[MAXN];Panax Notoginseng     intS; -CIN >> V >> E >>S; the      +      for(inti =1; I <= V; ++i) A          for(intj =1; J <= V; ++j) theRCOST[I][J] = cost[i][j] =INF; +      -      for(inti =0; i < E; ++i) { $Cin >> x >> y >>W; $RCOST[Y][X] = Cost[x][y] =W; -     } -      theDijkstra (S, cost, D);//calculate the shortest path separately - Dijkstra (S, rcost, RD);Wuyi      the      for(inti =1; I <= V; ++i)//sum -Sum[i] = D[i] +Rd[i]; Wu      -     intMaxnum = sum[1]; About      for(inti =1; I <= V; ++i)///find the maximum value $         if(Sum[i] >maxnum) -Maxnum =Sum[i]; -      -cout << Maxnum <<Endl; A      +     return 0; the}

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<poj-3268> Silver Cow Party Shortest Path problem