（POJ 1797） Heavy Transportation 最大生成树

题目链接：http://poj.org/problem?id=1797

Description

Background
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know. 

Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from  to n. Your task is to find the maximum weight that can be transported from crossing  (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.
Input

The first line contains the number of scenarios (city plans). For each city the number n of street crossings ( <= n <= ) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than . There will be at most one street between each pair of crossings.
Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at . Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.
Sample Input

Sample Output

Scenario #:

题目大意：有N个城市，有M条路，每条路上有一个最大承重量，问从1到N的道路上能通过的最大承重量是多少？

思路：就是求最大生成树上的最小值，dis【i】表示1到i的最大承重数

#include<stdio.h>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include <stack>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof(a))
#define N 1010
int Map[N][N];
int vis[N],dis[N],n,minn;
int dij(int s)
{
    vis[s]=;
    for(int i=;i<=n;i++)
        dis[i]=Map[s][i];
    for(int i=;i<n;i++)
    {
        int ans=-INF,k=;
        for(int j=;j<=n;j++)
        {
            if(!vis[j] && ans<dis[j])
                ans=dis[k=j];    /// 找到之中的最大值
        }
        vis[k]=;
        for(int j=;j<=n;j++)
        {
            if(!vis[j])
            {
                int m=min(Map[k][j],dis[k])  ///经过k点到j点，取从1到k点的最大承重量与从k到j点之间的最大承重量之间较小的值
            }
             dis[j]=max(dis[j],k);///从1到j是否要经过k点，如果经过k点的最大承重量大就经过k点
        }
    }
    return dis[n];///1到每个点的最大承重量
}
int main()
{
    int t,m,x,b,l,con=;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        met(Map,);
        for(int i=;i<m;i++)
        {
            scanf("%d %d %d",&x,&b,&l);
            Map[x][b]=Map[b][x]=l;          ///道路是双向的
        }
        met(vis,);
        printf("Scenario #%d:\n%d\n\n",con++,dij());
    }
    return ;
}