2017ccpc全国邀请赛(湖南湘潭) H. Highway

H. Highway
In ICPCCamp there were n towns conveniently numbered with 1,2,...,n connected with (n−1) roads. The
i-th road connecting towns a i and b i has length c i . It is guaranteed that any two cities reach each other
using only roads.
Bobo would like to build (n − 1) highways so that any two towns reach each using only highways. Building
a highway between towns x and y costs him δ(x,y) cents, where δ(x,y) is the length of the shortest path
between towns x and y using roads.
As Bobo is rich, he would like to find the most expensive way to build the (n − 1) highways.
Input
The input contains zero or more test cases and is terminated by end-of-file. For each test case:
The first line contains an integer n. The i-th of the following (n − 1) lines contains three integers a i , b i and
c i .
• 1 ≤ n ≤ 10 5
• 1 ≤ a i ,b i ≤ n
• 1 ≤ c i ≤ 10 8
• The number of test cases does not exceed 10.
Output
For each test case, output an integer which denotes the result.
Sample Input
5
1 2 2
1 3 1
2 4 2
3 5 1
5
1 2 2
1 4 1
3 4 1
4 5 2
Sample Output
19
15

官方题解：

先求出树的最远点对(树的直径)d1,d2，再求出以直径的两个端点为起点的dist[i](起点到i的距离)，首先将直径(d1,d2的距离)加入集合，对于其他点i，加入max(d1到i的距离,d2到i的距离)到集合，集合所构成的树就是题目的答案

至于树的最远点对怎么求就懒得说了

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define max(x,y) (x)>(y)?(x):(y)
#define min(x,y) (x)>(y)?(y):(x)
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1<<28;
const int maxn=100010;
const int maxm=110;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n;
struct node
{
    int next,to,cost;
}edge[maxn*2];
int head[maxn],cnt;
void add(int st,int ed,int cost)
{
    edge[cnt].to=ed;
    edge[cnt].cost=cost;
    edge[cnt].next=head[st];
    head[st]=cnt++;
}
ll distd1[maxn],distd2[maxn];
int vis[maxn];
void dfs(int x,ll d,ll dist[])
{
    vis[x]=1;
    dist[x]=d;
    for(int i=head[x];i!=-1;i=edge[i].next)
    {
        if(vis[edge[i].to]==0)
            dfs(edge[i].to,d+edge[i].cost,dist);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n))
    {
        cls(head,-1);
        cnt=0;
        for(int i=0;i<n-1;i++)
        {
            int a,b,c;
            scanf("%d %d %d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        cls(vis,0);
        dfs(1,0,distd1);
        int d1,d2;
        ll maxdist=0;
        for(int i=1;i<=n;i++)
            if(distd1[i]>maxdist)
                maxdist=distd1[i],d1=i;

        cls(vis,0);
        dfs(d1,0,distd1);
        maxdist=0;
        for(int i=1;i<=n;i++)
            if(distd1[i]>maxdist)
                maxdist=distd1[i],d2=i;
        cls(vis,0);
        dfs(d2,0,distd2);
        ll ans=distd1[d2];
        for(int i=1;i<=n;i++)
        {
            if(i!=d1&&i!=d2)
                ans+=max(distd1[i],distd2[i]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}