AYIT2017暑假集训第二周周三赛 B

The cows have run out of hay, a horrible event that must be remedied immediately. Bessie intends to visit the other farms to survey their hay situation. There are N (2 <= N <= 2,000) farms (numbered 1..N); Bessie starts at Farm 1. She'll traverse some or all of the M (1 <= M <= 10,000) two-way roads whose length does not exceed 1,000,000,000 that connect the farms. Some farms may be multiply connected with different length roads. All farms are connected one way or another to Farm 1. 

Bessie is trying to decide how large a waterskin she will need. She knows that she needs one ounce of water for each unit of length of a road. Since she can get more water at each farm, she's only concerned about the length of the longest road. Of course, she plans her route between farms such that she minimizes the amount of water she must carry. 

Help Bessie know the largest amount of water she will ever have to carry: what is the length of longest road she'll have to travel between any two farms, presuming she chooses routes that minimize that number? This means, of course, that she might backtrack over a road in order to minimize the length of the longest road she'll have to traverse.

Input

* Line 1: Two space-separated integers, N and M. 

* Lines 2..1+M: Line i+1 contains three space-separated integers, A_i, B_i, and L_i, describing a road from A_i to B_i of length L_i.

Output

* Line 1: A single integer that is the length of the longest road required to be traversed.

Sample Input

3 31 2 232 3 10001 3 43

Sample Output

43

Hint

OUTPUT DETAILS: 

In order to reach farm 2, Bessie travels along a road of length 23. To reach farm 3, Bessie travels along a road of length 43. With capacity 43, she can travel along these roads provided that she refills her tank to maximum capacity before she starts down a road.

题目的大概意思就是说给你几条边，让你组成一个最小生成树，挑出最小生成树里面的最大边

模板题目，我对自己没写出来也和无语

#include<stdio.h>#include<algorithm>using namespace std;struct note//存储顶点和边{    int x,y;    long long int z;//题目说长度很长，所以用long long int}l[50000];int n,m;int f[3000],sum,count1;//f存储根节点long long int cmp(note a,note b){    return a.z<b.z;}int gf(int v)//找根函数，确定自己和谁在一块{    if(f[v]==v)        return v;    else    {        f[v]=gf(f[v]);        return f[v];    }}int mer(int v,int u)//合并两子集合的函数{    int t1,t2;    t1=gf(v);    t2=gf(u);    if(t1!=t2)//判断两个节点是不是在同一个集合里面    {        f[t2]=t1;//进行合并        return 1;    }    return 0;}int main(){    int i;    while(~scanf("%d%d",&n,&m))    {        for(i=0;i<m;i++)            scanf("%d%d%lld",&l[i].x,&l[i].y,&l[i].z);        for(i=1;i<=n;i++)            f[i]=i;//对存储根的数组进行初始化        long long int k=0;        count1=0;        sort(l,l+m,cmp);//对其边长进行排序        //for(i=0;i<m;i++)           // printf("%d %d %lld\n",l[i].x,l[i].y,l[i].z);        for(i=0;i<m;i++)//开始对每一条边进行遍历        {            if(mer(l[i].x,l[i].y))//如果不在同一个集合里面，经用他            {                count1++;                k=max(k,l[i].z);//找出最小生成树里面那条最大的边            }            if(count1==n-1)break;        }        printf("%lld\n",k);    }}