描述 Description
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.
牛们干草要用完了！贝茜打算去勘查灾情．
有 N(2≤N≤2000) 个农场，M(≤M≤10000) 条双向道路连接着它们，长度不超过 109．每一个农场均与农场 1 连通．贝茜要走遍每一个农场．她每走一单位长的路，就要消耗一单位的水．从一个农场走到另一个农场，她就要带上数量上等于路长的水．请帮她确定最小的水箱容量．也就是说，确定某一种方案，使走遍所有农场通过的最长道路的长度最小，必要时她可以走回头路．
输入格式 InputFormat
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.
第 1 行输入两个整数 N 和 M; 接下来 M 行，每行输入三个整数，表示一条道路的起点终点和长度．
输出格式 OutputFormat
Line 1: A single integer that is the length of the longest road required to be traversed.
输出一个整数，表示在路线上最长道路的最小值．
样例输入 SampleInput
10 10
1 2 10
2 3 10
3 4 14
4 5 10
5 6 13
6 7 10
7 8 17
8 9 10
9 10 11
4 9 15
样例输出 SampleOutput
15
来源 Source
Silver
代码 Code
求最小生成树最大边。
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf=0x7fffffff;
struct road
{
int l,r,val;
bool operator < (const road &temp) const
{
return val<temp.val;
}
}a[10005];
int i,j,t,n,m,k,z,y,x,ans=0;
int fa[2005];
inline void read(int &x)
{
x=0;char ch=getchar();
while (ch<'0' || ch>'9') ch=getchar();
while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
}
int getfa(int s)
{
if (fa[s]==s) return fa[s];
else fa[s]=getfa(fa[s]);
return fa[s];
}
int main()
{
read(n);read(m);
for (i=1;i<=n;i++) fa[i]=i;
for (i=1;i<=m;i++) read(a[i].l),read(a[i].r),read(a[i].val);
sort(a+1,a+m+1);
for (i=1;i<=m;i++) if (getfa(a[i].l)!=getfa(a[i].r))
{
fa[getfa(a[i].l)]=a[i].r;
ans=a[i].val;
}
printf("%d",ans);
return 0;
}