AGC018：Tree and Hamilton Path（dfs & 树 哈密尔顿回路）

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:

NA1 B1 C1A2 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

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51 2 53 4 72 3 32 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

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82 8 81 5 14 8 22 5 43 8 66 8 92 7 12

Sample Output 2

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132

题意：给一棵树，要求找出树上最长的一条的哈密尔顿回路。

思路：按边考虑，计算每条边最多能贡献几次，发现只要围着树的重心对称跑，每条边都能充分利用，边i的贡献为它的较小子树节点数*2*边权值，但有一条边例外，假如树的重心只有一个，该重心相连的一条边贡献减一，自然是减最小那条边。当有两个重心，减去两重心相连的边即可。

# include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 1e5+30;int n, id=0, cnt=0, Next[maxn], in[maxn], out[maxn], son[maxn]={0}, zhong[maxn];int msize = 0x3f3f3f3f, mid;LL ans = 0;struct node{int e, w, next;}edge[maxn<<1];void add_edge(int u, int v, int w){    edge[cnt] = {v, w, Next[u]};    Next[u] = cnt++;    edge[cnt] = {u, w, Next[v]};    Next[v] = cnt++;}void dfs(int cur, int pre, int w){    in[cur] = ++id;    int imax = 0;    for(int i=Next[cur]; i!=-1; i=edge[i].next)    {        int v = edge[i].e;        if(v == pre) continue;        dfs(v, cur, edge[i].w);        son[cur] += son[v] + 1;        imax = max(imax, son[v]+1);    }    out[cur] = id;    int imin = min(out[cur]-in[cur]+1, n-out[cur]+in[cur]-1);    ans = ans + (LL)imin*w*2;    imax = max(imax, n-1-son[cur]);    zhong[cur] = imax;    if(imax < msize)    {        msize = imax;        mid = cur;    }}int main(){    memset(Next, -1, sizeof(Next));    int a, b, c;    scanf("%d",&n);    for(int i=1; i<n; ++i)    {        scanf("%d%d%d",&a,&b,&c);        add_edge(a, b, c);    }    dfs(1, 0, 0);    int com = 0x3f3f3f3f;    for(int i=Next[mid]; i!=-1; i=edge[i].next)    {        int v = edge[i].e;        com = min(com, edge[i].w);        if(zhong[v] == zhong[mid])        {            ans -= edge[i].w;            return 0*printf("%lld\n",ans);        }    }    printf("%lld\n",ans-com);    return 0;}