Codeforces 61D. Eternal Victory 트리의 특성

D. Eternal Victory
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 xi, yi and wi (1 ≤ xi, yi ≤ n, 0 ≤ wi ≤ 2 × 104). xi and yi are two ends of a road and wi 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

3
1 2 3
2 3 4

output

7

input

3
1 2 3
1 3 3

output

9

시작점부터 모든 노드에 접근합니다. 경계권과 최소를 요구합니다.기점에서 종점까지의 경로를 제외하고 나머지 각 변은 두 번 방문한 것을 알 수 있다. 그래서 문제는 기점에서 시작하는 가장 긴 경로를 찾는 것이다. 답은 변권과 *2-최장 경로이다.

#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;
}