CF1088F. Ehab and a weird weight formula

Solution

这题大概是个大力找性质题（莫名感觉学习文化课有利于找性质？！？）。

性质1：不难发现一个点比它权值小的有且仅有一个（最小权点除外）

性质2：我们把最小权点作为根，原树形成一个小根堆。

点和边都有贡献显然很难受。我们把点的贡献移到边上，于是边的贡献变成：
$$\lceil lo{g}_{2}dis(u,v)\rceil \times min(a_u,a_v)+a_u+a_v$$
设一个点$x$在原树上比它权值小的点编号为$i{d}_x$。

性质3：新树上一个点只会选择权值比它小的点作为$father$。（不然还不如选$i{d}_x$，贡献为$a_{i{d}_x}$+$a_x$）

性质4：因此设当前要选择$x$在新树上的$father$，发现只可能在原树上$i{d}_x$到根的路径上，也就是原树上$x$的父亲到根的路径上（不然不如取这个点和$x$的$LCA$）。

然后对于形如$\lceil log(dis(x,y))\rceil$的式子，显然是把$x$到根的链分成$O(log)$段，每段的$\lceil log(dis(x,y))\rceil$相等，由于小根堆的性质，我们只会取每一段的最上面的点。

于是倍增维护一下祖先，$O(logn)$跳链即可。

总时间复杂度$O(nlgn)$。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; }
template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondtypedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int, int> PR;
typedef vector<int> VI; const lod eps = 1e-9;
const lod pi = acos(-1);
const int oo = 1 << 30;
const ll loo = 1ll << 60;
const int mods = 998244353;
const int inv2 = (mods + 1) >> 1;
const int MAXN = 300005;
const int INF = 0x3f3f3f3f; //1061109567
/*--------------------------------------------------------------------*/namespace FastIO{constexpr int SIZE = (1 << 21) + 1;int num = 0, f;char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)inline void flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}inline void putc(char c) {*oS ++ = c;if (oS == oT) flush();}inline void getc(char &c) {for (c = gc(); !isdigit(c) && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); isdigit(c) ; c = gc()) st[++ n] = c;st[n + 1] = '\0';}template<class I>inline void read(I &x) {for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);x *= f;}template<class I>inline void print(I x) {if (x < 0) putc('-'), x = -x;if (!x) putc('0');while (x) que[++ num] = x % 10 + 48, x /= 10;while (num) putc(que[num --]);}struct Flusher_{~Flusher_(){flush();}} io_Flusher_;
}
using FastIO :: read;
using FastIO :: putc;
using FastIO :: reads;
using FastIO :: print;ll ans = 0;
vector<int> e[MAXN];
int dep[MAXN], Log[MAXN], a[MAXN], fa[MAXN][20], id, n;
void dfs(int x, int father) {dep[x] = dep[father] + 1, fa[x][0] = father;for (int i = 1; i <= Log[dep[x]] ; ++ i) fa[x][i] = fa[fa[x][i - 1]][i - 1];for (auto v : e[x]) {if (v == father) continue;dfs(v, x);}if (father) {ll mn = 1ll * (Log[dep[x]] + 1) * a[id] + a[id] + a[x];for (int i = 0; i <= Log[dep[x]] ; ++ i) upmin(mn, 1ll * min(a[x], a[fa[x][i]]) * i + a[x] + a[fa[x][i]]);ans += mn;}
}
signed main() {
#ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin);
#endifread(n);Log[1] = 0, dep[0] = -1, id = 1;for (int i = 2; i <= n ; ++ i) Log[i] = Log[i >> 1] + 1;for (int i = 1; i <= n ; ++ i) {read(a[i]);if (a[i] < a[id]) id = i;}for (int i = 1, u, v; i < n ; ++ i) read(u), read(v), e[u].PB(v), e[v].PB(u);dfs(id, 0);print(ans);return 0;
}