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=1010;
const int INF=0x3f3f3f3f;
int cost[MAXN][MAXN];
int lowcost[MAXN];
bool vis[MAXN];
int n ;

void Dijkstra(int beg)
{
    for(int i=1;i<=n;i++)
    {
        lowcost[i]=0;
        vis[i]=false;
    }
    lowcost[beg]=INF;
    for(int j=0;j<n;j++)
    {
        int k=-1;
        int Max=0;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&lowcost[i]>Max)
            {
                Max=lowcost[i];
                k=i;
            }
        if(k==-1)break;
        vis[k]=true;
        for(int i=1;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=0;
    while(T--)
    {
        iCase++;
        scanf("%d%d",&n,&m);
        memset(cost,0,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(1);
        printf("Scenario #%d:
",iCase);
        printf("%d

",lowcost[n]);
    }
    return 0;
}