CF809E Surprise me! 莫比乌斯反演、虚树、树形DP

简化题意：给出一棵$n$个点的树，编号为$1$到$n$，第$i$个点的点权为$a_i$，保证序列$a_i$是一个$1$到$n$的排列，求

\[ \frac{1}{n(n-1)} \sum\limits_{i=1}^n \sum\limits_{j=1}^n \varphi(a_ia_j) dist(i,j)\]

其中$dist(i,j)$为树上$i,j$两点的距离。

看到$\varphi$第一反应推式子

因为序列$a_i$是一个$1$到$n$的排列，设$t_i$表示点权为$i$的点的编号，那么原式等于$ \frac{1}{n(n-1)} \sum\limits_{i=1}^n \sum\limits_{j=1}^n \varphi(ij) dist(t_i,t_j)$

接下来化简$\sum\limits_{i=1}^n \sum\limits_{j=1}^n\varphi(ij) $

考虑$gcd(i,j)$在$\varphi(ij)$中的贡献，不难得到$\varphi(ij) = \frac{\varphi(i) \varphi(j) gcd(i,j)}{\varphi(gcd(i,j))}$

代入得$ \sum\limits_{i=1}^n \sum\limits_{j=1}^n \varphi(ij) =\sum\limits_{i=1}^n \sum\limits_{j=1}^n \frac{\varphi(i) \varphi(j) gcd(i,j)}{\varphi(gcd(i,j))}$

按照套路枚举$gcd$：$=\sum\limits_{d=1}^n \frac{d}{\varphi(d)} \sum\limits_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{d} \rfloor} \varphi(id) \varphi(jd) [gcd(i,j) == 1]=\sum\limits_{d=1}^n \frac{d}{\varphi(d)} \sum\limits_{i=1}^{\lfloor \frac{n}{d} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{d} \rfloor} \varphi(id) \varphi(jd) \sum\limits_{p | gcd(i,j)} \mu(p)$

将$p$移到前面：$=\sum\limits_{d=1}^n \sum\limits_{p=1}^{\lfloor \frac{n}{d} \rfloor} \frac{d \mu(p)}{\varphi(d)} \sum\limits_{i=1}^{\lfloor \frac{n}{dp} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{dp} \rfloor} \varphi(idp) \varphi(jdp)$

$dp$太难看了考虑枚举$T=dp$：$=\sum\limits_{T=1}^n \sum\limits_{d | T} \frac{d \mu(\frac{T}{d})}{\varphi(d)} \sum\limits_{i=1}^{\lfloor \frac{n}{T} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{T} \rfloor} \varphi(iT) \varphi(jT)$

然后将$dist(t_i,t_j)$代回来。注意这个时候$i,j$的意义发生了变化，我们应该要代入的是$dist(t_{iT} , t_{jT})$

所以我们需要求的是$=\sum\limits_{T=1}^n \sum\limits_{d | T} \frac{d \mu(\frac{T}{d})}{\varphi(d)} \sum\limits_{i=1}^{\lfloor \frac{n}{T} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{T} \rfloor} \varphi(iT) \varphi(jT) dist(t_{iT} , t_{jT})$

推到这里我们就可以做了，相对来说还是比较好推的……

对于$\sum\limits_{d | T} \frac{d \mu(\frac{T}{d})}{\varphi(d)}$，枚举倍数做到$O(nlogn)$预处理

对于$\sum\limits_{i=1}^{\lfloor \frac{n}{T} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{T} \rfloor} \varphi(iT) \varphi(jT) dist(t_{iT} , t_{jT})$，它等于$\sum\limits_{i=1}^{\lfloor \frac{n}{T} \rfloor}\sum\limits_{j=1}^{\lfloor \frac{n}{T} \rfloor} \varphi(iT) \varphi(jT) (dep_{t_{iT}} + dep_{t_{jT}} - 2 * dep_{LCA(t_{iT} , t_{jT})})$。我们把所有点权为$T$的倍数的点拿出来建立虚树进行树形DP，对于每一个点维护它子树中满足条件的点的$\sum \varphi(i) \times dep_i$与$\sum \varphi(i)$，在两个子树合并时计算答案即可。总复杂度约为$O(nlog^2n)$。

最后记录一些被坑的细节（大概只有我这种菜鸡才会犯……）

$1.$建立虚树的点不是编号为$T$的倍数的点，而是点权是$T$的倍数的点

$2.$$\frac{d}{\varphi(d)}$不一定是整数

$3.$写程序的时候要保持清醒……发现很多地方$i$打成$j$，$a_x$打成$x$之类的……

$4.$记得清空数组

#include<bits/stdc++.h>
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    char c = getchar();
    bool f = 0;
    while(!isdigit(c) && c != EOF){
        if(c == ‘-‘)
            f = 1;
        c = getchar();
    }
    if(c == EOF)
        exit(0);
    while(isdigit(c)){
        a = a * 10 + c - 48;
        c = getchar();
    }
    return f ? -a : a;
}

const int MAXN = 2e5 + 9 , MOD = 1e9 + 7;
int mu[MAXN] , phi[MAXN] , prime[MAXN] , cnt;
bool nprime[MAXN];
struct Edge{
    int end , upEd;
}Ed[MAXN << 1];
int N , cntEd , ts , top , cntT , cntST , cur , ans1 , ans;
int head[MAXN] , val[MAXN] , ind[MAXN];
int dfn[MAXN] , dep[MAXN] , fir[MAXN] , ST[21][MAXN << 1] , logg2[MAXN << 1];
int st[MAXN] , tree[MAXN] , dp1[MAXN] , dp2[MAXN] , q[MAXN];
vector < int > ch[MAXN];

inline void addEd(int a , int b){
    Ed[++cntEd].end = b;
    Ed[cntEd].upEd = head[a];
    head[a] = cntEd;
}

void input(){
    N = read();
    for(int i = 1 ; i <= N ; ++i)
        ind[val[i] = read()] = i;
    for(int i = 1 ; i < N ; ++i){
        int a = read() , b = read();
        addEd(a , b);
        addEd(b , a);
    }
}

inline void init(){
    phi[1] = mu[1] = 1;
    for(int i = 2 ; i <= N ; ++i){
        if(!nprime[i]){
            prime[++cnt] = i;
            phi[i] = i - 1;
            mu[i] = -1;
        }
        for(int j = 1 ; j <= cnt && prime[j] * i <= N ; ++j){
            nprime[i * prime[j]] = 1;
            if(i % prime[j] == 0){
                phi[i * prime[j]] = phi[i] * prime[j];
                break;
            }
            phi[i * prime[j]] = phi[i] * (prime[j] - 1);
            mu[i * prime[j]] = mu[i] * -1;
        }
    }
}

void init_dfs(int x , int p){
    dfn[x] = ++ts;
    ST[0][++cntST] = x;
    fir[x] = cntST;
    dep[x] = dep[p] + 1;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(Ed[i].end != p){
            init_dfs(Ed[i].end , x);
            ST[0][++cntST] = x;
        }
}

inline int cmp(int a , int b){
    return dep[a] < dep[b] ? a : b;
}

void init_ST(){
    for(int i = 2 ; i <= cntST ; ++i)
        logg2[i] = logg2[i >> 1] + 1;
    for(int i = 1 ; 1 << i <= cntST ; ++i)
        for(int j = 1 ; j + (1 << i) <= cntST + 1 ; ++j)
            ST[i][j] = cmp(ST[i - 1][j] , ST[i - 1][j + (1 << (i - 1))]);
}

inline int LCA(int x , int y){
    x = fir[x];
    y = fir[y];
    if(x > y)
        swap(x , y);
    int t = logg2[y - x + 1];
    return cmp(ST[t][x] , ST[t][y - (1 << t) + 1]);
}

inline int poww(long long a , int b){
    int times = 1;
    while(b){
        if(b & 1)
            times = times * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return times;
}

bool cmp1(int a , int b){
    return dfn[a] < dfn[b];
}

int dfs(int x){
    dp1[x] = dp2[x] = 0;
    int sum = 0;
    for(int i = 0 ; i < ch[x].size() ; ++i){
        sum = (sum + dfs(ch[x][i])) % MOD;
        sum = (sum + 1ll * dp1[x] * dp2[ch[x][i]] % MOD + 1ll * dp1[ch[x][i]] * dp2[x] % MOD - 2ll * dep[x] * dp1[x] % MOD * dp1[ch[x][i]] % MOD + MOD) % MOD;
        dp1[x] = (dp1[x] + dp1[ch[x][i]]) % MOD;
        dp2[x] = (dp2[x] + dp2[ch[x][i]]) % MOD;
    }
    if(val[x] % cur == 0){
        sum = (sum + 1ll * dp2[x] * phi[val[x]] % MOD - 1ll * dep[x] * dp1[x] % MOD * phi[val[x]] % MOD + MOD) % MOD;
        dp1[x] = (dp1[x] + phi[val[x]]) % MOD;
        dp2[x] = (dp2[x] + 1ll * phi[val[x]] * dep[x]) % MOD;
    }
    return sum;
}

inline void calc(int x){
    cntT = 0;
    for(int i = 1 ; x * i <= N ; ++i){
        ch[ind[x * i]].clear();
        tree[++cntT] = ind[x * i];
    }
    sort(tree + 1 , tree + cntT + 1 , cmp1);
    for(int i = 1 ; i <= cntT ; ++i){
        if(top){
            int t = LCA(st[top] , tree[i]);
            while(top - 1 && dep[st[top - 1]] >= dep[t]){
                ch[st[top - 1]].push_back(st[top]);
                --top;
            }
            if(t != st[top]){
                ch[t].clear();
                ch[t].push_back(st[top]);
                st[top] = t;
            }
        }
        st[++top] = tree[i];
    }
    int rt = st[1];
    while(top > 1){
        ch[st[top - 1]].push_back(st[top]);
        --top;
    }
    top = 0;
    ans = (ans + 1ll * q[x] * dfs(rt)) % MOD;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    input();
    init();
    init_dfs(1 , 0);
    init_ST();
    for(int i = 1 ; i <= N ; ++i)
        for(int j = 1 ; j * i <= N ; ++j)
            q[i * j] = (q[i * j] + 1ll * i * mu[j] * poww(phi[i] , MOD - 2) % MOD + MOD) % MOD;
    for(cur = 1 ; cur <= N / 2 ; ++cur)
        calc(cur);
    cout << 2ll * ans * poww(1ll * N * (N - 1) % MOD , MOD - 2) % MOD;
    return 0;
}

原文地址：https://www.cnblogs.com/Itst/p/10284347.html

时间： 2024-08-30 13:38:52

CF809E Surprise me! 莫比乌斯反演、虚树、树形DP

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