Highway

In ICPCCamp there were n towns conveniently numbered with 1,2,…,n connected with (n−1) roads. The i-th road connecting towns ai and bi has length ci. 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 ai, bi and ci.

1≤n≤105
1≤ai,bi≤n
1≤ci≤108
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
题目链接： http://202.197.224.59/OnlineJudge2/index.php/Problem/index/p/13/
或 https://www.icpc-camp.org/contests/4mYguiUR8k0GKE
题目大意：
给出一棵无根树，给出n-1条边和权值，然后让你求重新选择n-1条边使其联通并且权和最大
官方题解：

首先有一点得知道，关于树的直径
树的直径是指树上权值和最大的路径（最简单路径，即每一个点只经过一次）
存在结论：对于树上的任意一个节点，距离这个节点最远的距离一定是到直径的端点的距离

就是先求出树的最远点对(树的直径的端点)d1,d2，再求出以直径的两个端点为起点的disi，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

#include <iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<queue>
#include<cstdio>
using namespace std;
const int maxn = 100100;
struct node
{
    int u,v,w;
    int next;
} e[maxn<<1];
long long dis[maxn],dd[maxn];
int head[maxn];
int top;
bool vis[maxn];
void init()
{
    memset(dis,0,sizeof(dis));
    memset(head,-1,sizeof(head));
    memset(vis,false,sizeof(vis));
    top = 0;
}
void add(int u,int v,int w)
{
    e[top].u = u;
    e[top].v = v;
    e[top].w = w;
    e[top].next = head[u];
    head[u] = top++;

}
int bfs(int s)//求树的直径
{
    queue<int>q;
    vis[s] = true;
    q.push(s);
    int key = - 1; 
    long long mmax = 0;
    while(!q.empty())
    {
        int t = q.front();
        q.pop();


        for(int i=head[t]; ~i; i=e[i].next)
        {
            int v = e[i].v;//当前思想，把这个点看成结果理解下面的if
            if(!vis[v]&&dis[v]<dis[t]+e[i].w)//最大路判断
            {
                vis[v] = true;
                dis[v] = dis[t]+e[i].w;
                q.push(v);
                if(dis[v]>mmax)
                {
                    mmax = dis[v];
                    key = v;
                }
            }
        }

    }
    return key;
}
void bfs1(int s)
{
    queue<int>q;
    vis[s] = true;
    q.push(s);
    while(!q.empty())
    {
        int t = q.front();
        q.pop();
        for(int i=head[t]; ~i; i =e[i].next)
        {
            int v = e[i].v;
            if(!vis[v]&&dis[v]<dis[t]+e[i].w)
            {
                vis[v] =true;
                dis[v] = dis[t]+e[i].w;
                dd[v] = max(dis[v],dd[v]);
                q.push(v);
            }
        }
    }
}
int main()
{
    int n,u,v,w;
    while(cin>>n)
    {
        init();
        memset(dd,0,sizeof(dd));
        for(int i=0; i<n-1; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        int a = bfs(1);
        memset(vis,false,sizeof(vis));
        memset(dis,0,sizeof(dis));
        int b = bfs(a);
        long long sum = dis[b];
        memset(vis,false,sizeof(vis));
        memset(dis,0,sizeof(dis));
        bfs1(a);
        memset(vis,false,sizeof(vis));
        memset(dis,0,sizeof(dis));
        bfs1(b);
        for(int i=1; i<=n; i++)
        {
            if(i==a||i==b)
                continue;
            sum+=dd[i];
        }
        cout<<sum<<endl;
    }
    return 0;
}

参考：
http://blog.csdn.net/bless924295/article/details/72331272
http://blog.csdn.net/liangzhaoyang1/article/details/72796738