2017年“嘉杰信息杯” 中国大学生程序设计竞赛全国邀请赛 Highway

Highway
Accepted : 122		Submit : 393
Time Limit : 4000 MS		Memory Limit : 65536 KB

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≤10⁵
• 1≤a_i,b_i≤n
• 1≤c_i≤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

Source

XTU OnlineJudge 

 题意：要把所有的边都联通，并要求权值之和最大

 解法：因为是颗树，那就没必要讨论两点的最短路了（毕竟树是没有回路的），那么重点是落在了如何找到权值之和最大的方法

 我们找到这颗树相距最远的两个点（树的直径）A，B 剩余的点取距离A或者B最远的那条，需要进行两次dfs，距离保存最大的

 然后加起来就行，因为A到B我们多加了一次，再减去就是结果

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int inf=(1<<30);
const int maxn=100005;
ll pos;
ll n,ans,vis[maxn],in[maxn];
vector<pair<int,int>>e[maxn];
ll sum;
void dfs(int v,ll cnt)
{
    if(ans<cnt)
    {
        ans=cnt;
        pos=v;
    }
    if(vis[v])return;
    vis[v]=1;
    for(int i=0; i<e[v].size(); i++)
        //    cout<<e[v][i].first;
        if(!vis[e[v][i].first])
            dfs(e[v][i].first,cnt+e[v][i].second);
}
ll dis1[123456],dis2[123456];
void DFS(int v,ll cnt,ll dis[])
{
    if(vis[v]) return;
    vis[v]=1;
    dis[v]=cnt;
    for(int i=0; i<e[v].size(); i++)
        //    cout<<e[v][i].first;
        if(!vis[e[v][i].first])
            DFS(e[v][i].first,cnt+e[v][i].second,dis);
}
int main()
{
    int n,m;
    ans=0;
    while(~scanf("%d",&n))
    {
        ans=0;
        memset(dis1,0,sizeof(dis1));
        memset(dis2,0,sizeof(dis2));
        memset(in,0,sizeof(in));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<=n;i++)
        {
            e[i].clear();
        }
        for(int i=1; i<n; i++)
        {
            ll u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            e[u].push_back({v,w});
            e[v].push_back({u,w});
        }
        dfs(1,0);
        ll cnt=ans;
        ans=0;
        memset(vis,0,sizeof(vis));
        ans=0;
        DFS(pos,0,dis1);
        memset(vis,0,sizeof(vis));
        ans=0;
        dfs(pos,0);

        memset(vis,0,sizeof(vis));
        DFS(pos,0,dis2);
        memset(vis,0,sizeof(vis));
        ll cot=ans;
        //cout<<cot<<" "<<cnt<<endl;
        ll Max=max(cnt,cot);
        //cout<<Max<<endl;
        sum=0;
        for(int i=1;i<=n;i++)
        {
            sum+=max((ll)dis1[i],(ll)dis2[i]);
        }
        printf("%lld
",sum-Max);
    }
    return 0;
}