最小生成树最大边权（克鲁斯卡尔）

1.

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 Nis 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 Cis the length in kilometers between city Xand 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

#include 
#include 
#include 
#define Inf 0x7fffffff
using namespace std;
struct node
{
    int u,v;
    int w;
    node(int a,int b,int x):u(a),v(b),w(x){}
};
vector edge;
int vn,an;
int pa[100005];
bool cmp(const node &a,const node &b)
{
    return a.w<b.w;
}
int find(int x)
{
    return x==pa[x]?x:pa[x]=find(pa[x]);
}
int kruskal()
{
    int sumw=0,cnt=0;
    for(int i=1;i<=vn;i++) pa[i]=i;
    int Max = -1;
    for(int i=0;i)
    {
        int x=find(edge[i].u),y=find(edge[i].v);
        if(x!=y)
        {
            pa[x]=y;
            sumw+=edge[i].w;
            Max = max(edge[i].w, Max);
            ++cnt;
        }
        if(cnt==vn-1) break;
    }
    return Max;
}
int main()
{
    int T;
    cin>>T;
    while(T--){
        cin>>vn>>an;
        edge.clear();
        for(int i=1;i<=an;i++)
        {
            int a,b,x;
            cin>>a>>b>>x;
            edge.push_back(node(a,b,x));
        }
        sort(edge.begin(),edge.end(),cmp);
        cout<endl;
    }
    return 0;
}

 

Sample Output

4
2

题意：找到能在各个城市中穿行所需要的最小油量；

思路：找到图的最小生成树并找到其中最大的一条边权；

 

转载于:https://www.cnblogs.com/cccyx/p/11580638.html