（树的重心/DFS/构造）AtCoder Grand Contest 018 D

D - Tree and Hamilton Path

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Time limit : 2sec / Memory limit : 256MB

Score : 1100 points

Problem Statement

There is a tree with N vertices, numbered 1 through N. The i-th edge in this tree connects Vertices Ai and Bi and has a length of Ci.

Joisino created a complete graph with N vertices. The length of the edge connecting Vertices u and v in this graph, is equal to the shortest distance between Vertices u and v in the tree above.

Joisino would like to know the length of the longest Hamiltonian path (see Notes) in this complete graph. Find the length of that path.

Notes

A Hamiltonian path in a graph is a path in the graph that visits each vertex exactly once.

Constraints

• 2≤N≤105
• 1≤Ai<Bi≤N
• The given graph is a tree.
• 1≤Ci≤108
• All input values are integers.

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Input

Input is given from Standard Input in the following format:

N
A1 B1 C1
A2 B2 C2
:
AN−1 BN−1 CN−1

Output

Print the length of the longest Hamiltonian path in the complete graph created by Joisino.

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Sample Input 1

Copy

5
1 2 5
3 4 7
2 3 3
2 5 2

Sample Output 1

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38

The length of the Hamiltonian path 5 → 3 → 1 → 4 → 2 is 5+8+15+10=38. Since there is no Hamiltonian path with length 39 or greater in the graph, the answer is 38.

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Sample Input 2

Copy

8
2 8 8
1 5 1
4 8 2
2 5 4
3 8 6
6 8 9
2 7 12

Sample Output 2

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132

依然是按照官方题解的做法写的。DFS的方法比较巧妙，c数组 按照DFS的“方向”，记录某一节点在该方向上（包括自身）所有后代节点个数。这样统计官方题解中的si就非常方便了。

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <vector>
#include <stack>
#define mp make_pair
#define MIN(a,b) (a>b?b:a)
//#define MAX(a,b) (a>b?a:b)
typedef long long ll;
typedef unsigned long long ull;
const int MAX=1e5+5;
const int INF=1e9+5;
using namespace std;
//const int MOD=1e9+7;
typedef pair<ll,int> pii;
typedef pair<ll,ll> pll;
const double eps=0.00000001;
ll n,an,m=INF;
vector<pll>edge[MAX];
ll c[MAX];
void calc(ll now,ll pre)
{
    c[now]=1;
    for(auto &X:edge[now])
    {
        ll to,val;
        tie(to,val)=X;
        if(to==pre)
            continue;
        calc(to,now);
        c[now]+=c[to];
    }
}
void solve(ll now,ll pre,ll len)//从pre连到now的一条长为len的边
{
    an+=2*len*min(c[now],n-c[now]);
    pll mx={-1,-1};//now的最大子树的节点数 及连向其的边的权值
    for(auto &X:edge[now])
    {
        ll to,val;
        tie(to,val)=X;
        if(to==pre)
            continue;
        mx=max(mx,{c[to],val});
    }
    mx=max(mx,{n-c[now],len});
    if(2*mx.first==n)
        m=min(m,mx.second);
    else if(2*mx.first<n)
    {
        for(auto &X:edge[now])
            m=min(m,X.second);
    }
    for(auto &X:edge[now])
    {
        ll to,val;
        tie(to,val)=X;
        if(to==pre)
            continue;
        solve(to,now,val);
    }
}
 
int main()
{
    scanf("%lld",&n);
    for(ll i=1;i<n;i++)
    {
        ll A,B,C;
        scanf("%lld%lld%lld",&A,&B,&C);
        edge[A].push_back({B,C});
        edge[B].push_back({A,C});
    }
    calc(1,0);
    solve(1,0,0);
    printf("%lld
",an-m);
 
}