Highway

Highway
Accepted : 78		Submit : 275
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 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

//题意:有一棵树，问所有点到这棵树中所有点最远距离和为多少

//因为这个是树，所以，从任意一点 dfs 搜最远点，再从最远点 dfs 搜最远点，再dfs一下，这样，保存了2个最远点到任意点的距离，遍历一遍选最大的那个即可，最后减去直径，因为重复算了一次

一共就 4 n 次遍历吧，1800 ms 还行吧

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 #include <vector>
 #include <queue>
 using namespace std;
 #define LL long long
 #define INF 0x3f3f3f3f3f3f3f3f
 #define MX 100005
 struct To
 {
     int to;
     LL c;
 };

 int n;
 int far;
 LL step;
 vector<To> road[MX];
 LL dis[][MX];

 void Init()
 {
     for (int i=;i<=n;i++)
         road[i].clear();
 }

 void dfs(int x,int pre,LL s,int kk)//所在位置，从前，距离，第几次
 {
     if (kk!=) dis[kk-][x]=s;

     if (s>step)
     {
         far=x;
         step=s;
     }
     for (int i=;i<(int)road[x].size();i++)
     {
         if (road[x][i].to!=pre)
             dfs(road[x][i].to,x,road[x][i].c+s,kk);
     }
 }

 int main()
 {
     while (~scanf("%d",&n))
     {
         Init();
         for (int i=;i<n-;i++)
         {
             int a,b,c;
             scanf("%d%d%d",&a,&b,&c);
             road[a].push_back((To){b,c});
             road[b].push_back((To){a,c});
         }

         step=;
         dfs(,,,);

         step=;
         dfs(far,,,);
         step=;
         dfs(far,,,);

         LL ans = ;
         for (int i=;i<=n;i++)
             ans += max(dis[][i],dis[][i]);
         ans -= step;
         printf("%I64d\n",ans);
     }
     return ;
 }