Codeforces 61D. Eternal Victory 树的性质

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Valerian was captured by Shapur. The victory was such a great one that Shapur decided to carve a scene of Valerian's defeat on a mountain. So he had to find the best place to make his victory eternal!

He decided to visit all n cities of Persia to find the best available mountain, but after the recent war he was too tired and didn't want to traverse a lot. So he wanted to visit each of these n cities at least once with smallest possible traverse. Persian cities are connected with bidirectional roads. You can go from any city to any other one using these roads and there is a unique path between each two cities.

All cities are numbered 1 to n. Shapur is currently in the city 1 and he wants to visit all other cities with minimum possible traverse. He can finish his travels in any city.

Help Shapur find how much He should travel.

Input

First line contains a single natural number n (1 ≤ n ≤ 10⁵) — the amount of cities.

Next n - 1 lines contain 3 integer numbers each x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n, 0 ≤ w_i ≤ 2 × 10⁴). x_i and y_i are two ends of a road and w_i is the length of that road.

Output

A single integer number, the minimal length of Shapur's travel.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample test(s)
input
31 2 32 3 4
output
7
input
31 2 31 3 3
output
9

从起点开始访问所有的节点，要求边权和最小。可以发现除了从起点到终点一条路径，剩余每条边都访问了两次，所以问题变成了寻找从起点开始的最长路径，答案就是边权和*2-最长路径。

#include <iostream>#include <cstring>#include <cstdio>using namespace std;const int INF=1e9+9;const int maxn=111111;int n;long long ans;struct EDGE{    int to;    int w;    int next;}edges[maxn*2];int edge;int head[maxn];void init(){    memset(head,-1,sizeof(head));    memset(edges,0,sizeof(edges));    edge=0;    ans=0;}void addedge(int u,int v,int c){    edges[edge].w=c;edges[edge].to=v;edges[edge].next=head[u];head[u]=edge++;    edges[edge].w=c;edges[edge].to=u;edges[edge].next=head[v];head[v]=edge++;}long long dfs(int u,int dad,long long flow){    long long ret=flow;    int v,w;    for (int i=head[u];i!=-1;i=edges[i].next)    {        v=edges[i].to;        w=edges[i].w;        if (v!=dad)        {            ret=max( ret, dfs(v,u,flow+w) );        }    }    return ret;}int main(){    int n;    cin>>n;    init();    for (int i=1;i<n;i++)    {        int u,v,w;        cin>>u>>v>>w;        addedge(u,v,w);        ans+=w*2;    }    long long ret=dfs(1,0,0);    cout<<ans-ret<<endl;    return 0;}