Little X has a tree consisting of n nodes (they are numbered from 1 to n). Each edge of the tree has a positive length. Let‘s define the distance between two nodes v and u (we‘ll denote it d(v, u)) as the sum of the lengths of edges in the shortest path between v and u.
A permutation p is a sequence of n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n). Little X wants to find a permutation p such that sum 
is maximal possible. If there are multiple optimal permutations, he wants to find the lexicographically smallest one. Help him with the task!
The first line contains an integer n (1 ≤ n ≤ 105).
Each of the next n - 1 lines contains three space separated integers ui, vi, wi (1 ≤ ui, vi ≤ n; 1 ≤ wi ≤ 105), denoting an edge between nodes ui and vi with length equal to wi.
It is guaranteed that these edges form a tree.
In the first line print the maximum possible value of the described sum. In the second line print n integers, representing the lexicographically smallest permutation.
2
1 2 3
6
2 1
5
1 2 2
1 3 3
2 4 4
2 5 5
32
2 1 4 5 3
写了一上午，我太辣鸡了。
∑dis(i,pi?)=∑(depi?+deppi??−2×deplca(i,pi?)?)=2×∑depi?−2×∑deplca(i,pi?)?
前面一定是不变的， 后面我们可以把它降低到0， 我们选择树的重心作为根，可以把后边变成0.
第一问解决。把一个点看成两个， 一个是入点，一个是出点.
记为in[x], out[x];要满足子树x可以匹配到答案， 我们必须让 $\large in[x] + out[x] <= \sum in[i]$
$in[x]$
$ f[i, j] = $
$\LARGE f[i, j]=\frac{\sum_{to}^{ } f[nxt[nxt[i,j],j],to] + f[nxt[nxt[i,j],j],j]}{deg[j]+1} + 1$
latex炸了不写了
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <set> using namespace std; inline int read() { int res = 0;char ch=getchar(); while(!isdigit(ch)) ch=getchar(); while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48), ch=getchar(); return res; } #define reg register #define N 100005 #define ll long long int n; struct edge { int nxt, to, val; }ed[N*2];int head[N], cnt; inline void add(int x, int y, int z) { ed[++cnt] = (edge) {head[x], y, z}; head[x] = cnt; } ll ans; int siz[N], root, mrt = 1e9; void dfs(int x, int fa) { siz[x] = 1; int tmp = 0; for (reg int i = head[x] ; i ; i = ed[i].nxt) { int to = ed[i].to; if (to == fa) continue; dfs(to, x); siz[x] += siz[to]; tmp = max(tmp, siz[to]); } tmp = max(tmp, n - siz[x]); if (tmp < mrt) mrt = tmp, root = x; } void gfs(int x, int fa, ll d) { ans += d; for (reg int i = head[x] ; i ; i = ed[i].nxt) { int to = ed[i].to; if (to == fa) continue; gfs(to, x, d + ed[i].val); } } set <int> all; set <int> tr[N]; set < pair <int, int> > inout; #define makp(a, b) make_pair(a, b) int belong[N], sum[N]; void ffs(int x, int fa, int bel) { belong[x] = bel; tr[bel].insert(x); siz[x] = 1; for (reg int i = head[x] ; i ; i = ed[i].nxt) { int to = ed[i].to; if (to == fa) continue; ffs(to, x, bel); siz[x] += siz[to]; } } inline int Find(int x) { set < pair <int, int> > :: iterator it = inout.lower_bound(makp(n - x + 1, 0)); bool must = 0; if (it != inout.end() and (*it).first == n - x + 1 and (*it).second != belong[x]) must = 1; if (belong[x]) { inout.erase(makp(sum[belong[x]], belong[x])); sum[belong[x]]--; inout.insert(makp(sum[belong[x]], belong[x])); } if (must) { int k = *tr[(*it).second].begin(); all.erase(k); tr[(*it).second].erase(k); if (tr[(*it).second].size()) all.insert(*tr[(*it).second].begin()); inout.erase(makp(sum[belong[k]], belong[k])); inout.insert(makp(--sum[belong[k]], belong[k])); return k; } else { set <int> :: iterator itt = all.begin(); if(belong[*itt] == belong[x] and x != root) itt++; all.erase(itt); int k = *itt; if (belong[k]) { tr[belong[k]].erase(k); if (tr[belong[k]].size()) all.insert(*tr[belong[k]].begin()); inout.erase(makp(sum[belong[k]], belong[k])); inout.insert(makp(--sum[belong[k]], belong[k])); } return k; } return 0; } int main() { n = read(); for (reg int i = 1 ; i < n ; i ++) { int x = read(), y = read(), z = read(); add(x, y, z), add(y, x, z); } root = 1; dfs(1, 0); ans = 0; gfs(root, 0, 0); printf("%lld\n", ans * 2); all.insert(root); for (reg int i = head[root] ; i ; i = ed[i].nxt) { int to = ed[i].to; ffs(to, root, to); sum[to] = siz[to] * 2; inout.insert(makp(sum[to], to)); siz[root] += siz[to]; all.insert(*tr[to].begin()); } for (reg int i = 1 ; i <= n ; i ++) printf("%d ", Find(i)); return 0; }
原文：https://www.cnblogs.com/BriMon/p/9547292.html