poj 1797 一条路径中的最小边 再找出最大的

Sample Input

1 // T
3 3// n m
1 2 3//u v w
1 3 4
2 3 5
Sample Output

Scenario #1:
4

 # include <iostream>
 # include <cstdio>
 # include <cstring>
 # include <algorithm>
 # include <cmath>
 # define LL long long
 using namespace std ;

 const int MAXN=;
 const int INF=0x3f3f3f3f;
 int cost[MAXN][MAXN];
 int lowcost[MAXN];
 bool vis[MAXN];
 int n ;

 void Dijkstra(int beg)
 {
     for(int i=;i<=n;i++)
     {
         lowcost[i]=;
         vis[i]=false;
     }
     lowcost[beg]=INF;
     for(int j=;j<n;j++)
     {
         int k=-;
         int Max=;
         for(int i=;i<=n;i++)
             if(!vis[i]&&lowcost[i]>Max)
             {
                 Max=lowcost[i];
                 k=i;
             }
         if(k==-)break;
         vis[k]=true;
         for(int i=;i<=n;i++)
             if(!vis[i]&&min(lowcost[k],cost[k][i])>lowcost[i] )
                 lowcost[i]=min(lowcost[k],cost[k][i]);
     }
 }

 int main()
 {
   //  freopen("in.txt","r",stdin) ;
     int T;
     int m;
     scanf("%d",&T);
     int iCase=;
     while(T--)
     {
         iCase++;
         scanf("%d%d",&n,&m);
         memset(cost,,sizeof(cost));
         int u,v,w;
         while(m--)
         {
             scanf("%d%d%d",&u,&v,&w);
             cost[u][v]=cost[v][u]=max(cost[u][v],w);
         }
         Dijkstra();
         printf("Scenario #%d:\n",iCase);
         printf("%d\n\n",lowcost[n]);
     }
     return ;
 }