CodeForces 375D :Tree and Queries dsu on tree 换根DP

传送门

题意

有n个数，你可以对每个数进行两种操作：

将一个数乘以2
将一个数变为其二分之一并向下取整

问最少通过多少次操作可以将这些数的值变为全部相同。

分析

好像很多题解都是写的暴力啊，我来补一个换根$DP$
考虑建树，$i$和$i\ast 2$之间连一条单向边，$i/2$和$i$之间连一条单向边，然后如果要把所有的数全部转换成1的话，就是记录每一个点的代表的数出现的次数乘以这个点的深度的和，然后要统计每一个点作为答案的情况，换一下根即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],e[M],ne[M],idx;
ll w[N],dep[N];
ll ans[N];
ll sz[N];
int n;

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u,int fa){
    sz[u] = w[u];
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        dep[j] = dep[u] + 1;
        dfs(j,u);
        sz[u] += sz[j];
    }
}

void dfs_(int u,int fa){
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j % 2 == 1 && sz[j] != n) ans[j] = INF;
        else ans[j] = ans[u] - sz[j] + (n - sz[j]);
        dfs_(j,u);
    }
}

int main() {
    read(n);
    memset(h,-1,sizeof h);
    for(int i = 1;i <= n;i++) {
        int x;
        read(x);
        w[x]++;
    }
    for(int i = 1;i <= 100000;i++){
        if(i * 2 <= 100000) add(i,i * 2);
        if(i / 2 > 0 && i % 2 == 1) add(i / 2,i);
    }
    dfs(1,0);
    for(int i = 2;i <= 100000;i++) {
        ans[1] +=  dep[i] * w[i];
    }
    dfs_(1,0);
    ll res = INF;
    for(int i = 1;i <= 100000;i++) res = min(res,ans[i]);
    dl(res);
    return 0;
}