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[2][MX];

void Init()
{
    for (int i=1;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!=0) dis[kk-1][x]=s;

    if (s>step)
    {
        far=x;
        step=s;
    }
    for (int i=0;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=0;i<n-1;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=0;
        dfs(1,0,0,0);

        step=0;
        dfs(far,0,0,1);
        step=0;
        dfs(far,0,0,2);

        LL ans = 0;
        for (int i=1;i<=n;i++)
            ans += max(dis[0][i],dis[1][i]);
        ans -= step;
        printf("%I64d
",ans);
    }
    return 0;
}