NTT模板 神奇的迷宫 NTT加点分治

Posted 2022-11-05 goto_1600

链接
题意就是求给定一棵树
求${\sum }_0^{n-1}{W}_l\sum \sum [dist(i,j)==L]\ast a[i]\ast a[j]$
大小为1e5
思路：
树上路径考虑点分治，顺便复习了一下点分治，就是每次递归选一个重心，总共期望是$logn$层，每一层我们可以用$nlogn$的复杂度。我们对每一种长度的路径考虑，只要满足i+j==L那么不难想出卷积，每一层递归在新根的作用下，维护A和B数组的卷积，然后贡献给答案那么最后答案就是${\sum }_1^{n-1}cnt[i]\ast w[i]\ast 2+{\sum }_1^{n}w[0]\ast a[i]\ast a[i]$,为啥要乘2，因为两个地点不同，有顺序，对于单点的情况，不用乘2。顺便偷一个NTT的板子玩玩

#include<bits/stdc++.h>
using namespace std;
#define int long long 
typedef long long ll;
const int mod=998244353;
const int MOD=mod;
const int MAXN=1000010;
const int N=1000010;
const int maxn = 5e6+7;
const int G = 3;
bool st[N];
vector<int>v[N];
int A[maxn],B[maxn];
int a[N];
int pt=0;
int ans=0;
int qt=0;
int W[N];
int maxlen=0;
int C[N];
int cnt[N];
ll qpow(ll a,ll b)

    ll ans = 1;
    while(b>0)
    
    	if(b&1) ans = ans*a%mod;
    	b>>=1;
    	a = a*a%mod;
	
	return ans%mod;

int len,r[MAXN];
ll x[MAXN],y[MAXN],w[MAXN];
 
inline ll q_pow(ll x,ll y)
    ll res = 1;
    while(y)
        if(y & 1) res = res * x % MOD;
        x = x * x % MOD;
        y >>= 1;
    
    return res;

void NTT(ll *a,ll f)
    for(int i = 0;i < len;i ++)
        if(i < r[i]) swap(a[i],a[r[i]]);
    
    w[0] = 1;
    for(int i = 2;i <= len;i *= 2)
        ll wn;
        if(f == 1) wn = q_pow(G,(ll)(MOD-1)/i);
        else wn = q_pow(G,(ll)(MOD-1)-(MOD-1)/i);
        for(int j = i/2;j >= 0;j -= 2) w[j] = w[j/2];
        for(int j = 1;j < i/2;j += 2) w[j] = (w[j-1]*wn)%MOD;
        for(int j = 0;j < len;j += i)
            for(int k = 0;k < i/2;k ++)
                ll u = a[j+k],v = (a[j+k+i/2] * w[k]) % MOD;
                a[j+k] = (u + v) % MOD;
                a[j+k+i/2] = (u - v + MOD) % MOD;
            
        
    
    if(f == -1)
        ll inv = q_pow(len,MOD-2);
        for(int i = 0;i < len;i ++) a[i] = a[i] * inv % MOD;
    

 
void MUL(ll *a,ll *b,ll *c,ll n,ll m)
    len = 1;
    while(len <= (n + m)) len *= 2;
    int k = trunc(log(len + 0.5) / log(2));
    for(int i = 0;i < len;i ++)
        r[i] = (r[i>>1]>>1) | ((i&1) << (k-1));
    
    for(int i = 0;i < len;i ++)
        if(i < n) x[i] = a[i];else x[i] = 0;
        if(i < m) y[i] = b[i];else y[i] = 0;
    
    NTT(x,1);
    NTT(y,1);
    for(int i = 0;i < len;i ++) c[i] = x[i] * y[i] % MOD;
    NTT(c,-1);

 
int get_wc(int u,int fa,int tot,int &rt)

	if(st[u])	return 0;
	int maxv=0;
	int sum=1;
	for(auto j:v[u])
		if(j==fa)	continue;
		int tt=get_wc(j,u,tot,rt);
		sum+=tt;
		maxv=max(maxv,tt);
	
	maxv=max(maxv,tot-sum);
	if(maxv<=tot/2)
		rt=u;
	return sum;

int get_sz(int u,int fa)
	if(st[u])	return 0;
	int res=1;
	for(auto j:v[u])
		if(j==fa)	continue;
		res+=get_sz(j,u);
	
	return res;


int mxlen=0;

void get_dist(int u,int fa,int dist)

	if(st[u])	return ;
	mxlen=max(mxlen,dist);
	B[dist]=(B[dist] + a[u])%mod;
	for(auto j:v[u])
	
		if(j!=fa)
			get_dist(j,u,dist+1);
	

void dfs(int u,int fa)
	if(st[u])	return ;
	get_wc(u,fa,get_sz(u,fa),u);
	st[u]=true;
	maxlen=0;
	A[0]=a[u];
	for(auto j:v[u])
	
		if(j==fa)	continue;
		mxlen=0;
		get_dist(j,-1,1);
		MUL(A,B,C,maxlen+1,mxlen+1);	
		for(int k=0;k<maxlen+2+mxlen;k++)
			cnt[k]=(cnt[k]+C[k])%mod;
		for(int k=0;k<=mxlen;k++)
		
			A[k]=(A[k]+B[k])%mod;
			B[k]=0;
		
		maxlen=max(maxlen,mxlen);
	
	for(int k=0;k<=maxlen;k++)
	
		A[k]=0;
	
	for(auto j:v[u])	dfs(j,u);

signed main()

	int n; 
	cin>>n;
	int sum=0;
	for(int i=1;i<=n;i++)	cin>>a[i],sum+=a[i],sum%=mod;
	for(int i=1;i<=n;i++)	a[i]=a[i] * qpow(sum,mod-2)%mod;
	for(int i=0;i<=n-1;i++)
		cin>>W[i];
	for(int i=1;i<=n;i++)
	
		ans=(ans+a[i]*a[i]%mod*W[0])%mod;
	
	for(int i=0;i<n-1;i++)
	
		int a,b;
		cin>>a>>b;
		v[a].push_back(b);
		v[b].push_back(a);
		
	dfs(1,-1);
	for(int i=1;i<=n-1;i++)
	
		ans=(ans+2 * cnt[i]*W[i])%mod;
	
	cout<<ans<<endl;
	return 0;