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=(<<);
 const int maxn=;
 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]=;
     for(int i=; 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[],dis2[];
 void DFS(int v,ll cnt,ll dis[])
 {
     if(vis[v]) return;
     vis[v]=;
     dis[v]=cnt;
     for(int i=; 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=;
     while(~scanf("%d",&n))
     {
         ans=;
         memset(dis1,,sizeof(dis1));
         memset(dis2,,sizeof(dis2));
         memset(in,,sizeof(in));
         memset(vis,,sizeof(vis));
         for(int i=;i<=n;i++)
         {
             e[i].clear();
         }
         for(int i=; 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(,);
         ll cnt=ans;
         ans=;
         memset(vis,,sizeof(vis));
         ans=;
         DFS(pos,,dis1);
         memset(vis,,sizeof(vis));
         ans=;
         dfs(pos,);

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