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

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <cmath>
 6 # define LL long long
 7 using namespace std ;
 8
 9 const int MAXN=1010;
10 const int INF=0x3f3f3f3f;
11 int cost[MAXN][MAXN];
12 int lowcost[MAXN];
13 bool vis[MAXN];
14 int n ;
15
16 void Dijkstra(int beg)
17 {
18     for(int i=1;i<=n;i++)
19     {
20         lowcost[i]=0;
21         vis[i]=false;
22     }
23     lowcost[beg]=INF;
24     for(int j=0;j<n;j++)
25     {
26         int k=-1;
27         int Max=0;
28         for(int i=1;i<=n;i++)
29             if(!vis[i]&&lowcost[i]>Max)
30             {
31                 Max=lowcost[i];
32                 k=i;
33             }
34         if(k==-1)break;
35         vis[k]=true;
36         for(int i=1;i<=n;i++)
37             if(!vis[i]&&min(lowcost[k],cost[k][i])>lowcost[i] )
38                 lowcost[i]=min(lowcost[k],cost[k][i]);
39     }
40 }
41
42 int main()
43 {
44   //  freopen("in.txt","r",stdin) ;
45     int T;
46     int m;
47     scanf("%d",&T);
48     int iCase=0;
49     while(T--)
50     {
51         iCase++;
52         scanf("%d%d",&n,&m);
53         memset(cost,0,sizeof(cost));
54         int u,v,w;
55         while(m--)
56         {
57             scanf("%d%d%d",&u,&v,&w);
58             cost[u][v]=cost[v][u]=max(cost[u][v],w);
59         }
60         Dijkstra(1);
61         printf("Scenario #%d:\n",iCase);
62         printf("%d\n\n",lowcost[n]);
63     }
64     return 0;
65 }