Codeforces Travelling Salesman 最小生成树

Travelling Salesman
After leaving Yemen, Bahosain now works as a salesman in Jordan. He spends most of his time travelling
between different cities. He decided to buy a new car to help him in his job, but he has to decide about the
capacity of the fuel tank. The new car consumes one liter of fuel for each kilometer.
Each city has at least one gas station where Bahosain can refill the tank, but there are no stations on the roads
between cities.
Given the description of cities and the roads between them, find the minimum capacity for the fuel tank
needed so that Bahosain can travel between any pair of cities in at least one way.
Input
The first line of input contains T (1 ≤ T ≤ 64) that represents the number of test cases.
The first line of each test case contains two integers: N (3 ≤ N ≤ 100,000) and M (N-1 ≤ M ≤ 100,000) ,
where N is the number of cities, and M is the number of roads.
Each of the following M lines contains three integers: X Y C (1 ≤ X, Y ≤ N)(X ≠ Y)(1 ≤ C ≤ 100,000) , where
C is the length in kilometers between city X and city Y . Roads can be used in both ways.
It is guaranteed that each pair of cities is connected by at most one road, and one can travel between any pair
of cities using the given roads.
Output
For each test case, print a single line with the minimum needed capacity for the fuel tank.

Sample Input 
2
6 7
1 2 3
2 3 3
3 1 5
3 4 4
4 5 4
4 6 3
6 5 5
3 3
1 2 1
2 3 2
3 1 3

Sample Output

4
2

 

题意：城市里才有加油站，每一个城市都要去求油箱最小容量

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;

int t,n,m;
int father[100005];

struct node
{
    int st,ed,cos;
}a[100005];

int cmp(node a,node b)
{
    return a.cos<b.cos;
}

void Init(int x)
{
    for(int i=1;i<=x;i++)
        {
            father[i]=i;
        }
}

int Find(int x)
{
    if(x!=father[x])
    {
        father[x]=Find(father[x]);
    }
    return father[x];
}

int Union(int x,int y)
{
    x=Find(x);
    y=Find(y);
            //cout<<x<<"   fff   "<<y<<endl;

    if(x==y)
        return 0;
    father[x]=y;
    return 1;
}

int main()
{
    //freopen("input.txt","r",stdin);
    scanf("%d",&t);
    while(t--)
    {
          int res=0;
          scanf("%d%d",&n,&m);
          Init(n);
          for(int i=1;i<=m;i++)
            scanf("%d%d%d",&a[i].st,&a[i].ed,&a[i].cos);

          sort(a+1,a+1+m,cmp);
          for(int i=1;i<=m;i++)
          {
                  if(Union(father[a[i].st],father[a[i].ed]))
                  res=max(res,a[i].cos);
          }
          printf("%d\n",res);
    }
}