【LOJ】#6289. 花朵

题解

我当时连(n^2)的树背包都搞不明白，这道题稳稳的爆零啊= =

然后听说这道题需要FFT……我当时FFT的板子都敲不对，然后这道题就扔了

然后，我去考了thusc……好吧，令人不愉快的经历，听说我要是把这道题做了我大概就能A了D2T2……生无可恋.jpg

还有一个月，加油吧，NOI2018可能是我最后能去thu的机会了TAT

设dp[u][0 / 1][i]为以u为根的子树，没选u还是选了u，一共选了i个点
转移就是从所有子树里选出大小为i的独立集更新，转移可以类似树背包
这道题dp方程写出来卷积优化就是显然的，关键是怎么优化

我们把这个树给树链剖分了，设g[u][0 / 1][i]为u这个点除了u的重儿子以外的子树，没选u还是选了u，独立集大小为i的值（把i当成指数，把这个数组当成一个多项式）这是我们用来卷积的多项式

我们从深度最深的链开始，由于我们希望一下子算出一条链，汇总到链顶，而不关心链上每个点的dp值，用分治FFT把一条链的答案算出来，具体就是存四个多项式，记录这条链的头尾选或没选，然后合并起来

合并到父亲的时候我们对于每个点的所有轻儿子也分治乘起来，如果一个个乘起来会达到(n ^ 2)

分治一个链的复杂度是(O(size(p) log^{2} size(p)))p是重链顶端，然后因为轻重链剖分，所以(sum size(p) = O(N log N))复杂度为(O(N log^3 N))

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 80005
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 998244353,L = (1 << 18);
int W[L + 5],N,M,B[MAXN];
int siz[MAXN],dep[MAXN],fa[MAXN],son[MAXN];
int top[MAXN],Line[MAXN],tot,cnt,lsiz[MAXN],dfn[MAXN];
vector<int> f[MAXN][2],zero,g[2][MAXN];
struct node {
    int to,next;
}E[MAXN * 2];
struct res_node {
    vector<int> f00,f01,f10,f11;
};
int head[MAXN],sumE;
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
int inc(int a,int b) {
    a = a + b;
    if(a >= MOD) a -= MOD;
    return a;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
void NTT(vector<int> &a,int len,int on) {
    a.resize(len);
    for(int i = 1 , j = len / 2 ; i < len - 1 ; ++i) {
	if(i < j) swap(a[i],a[j]);
	int k = len / 2;
	while(j >= k) {
	    j -= k;
	    k >>= 1;
	}
	j += k;
    }
    for(int h = 2 ; h <= len ; h <<= 1) {
	int wn = W[(L + on * L / h) % L];
	for(int k = 0 ; k < len ; k += h) {
	    int w = 1;
	    for(int j = k ; j < k + h / 2 ; ++j) {
		int u = a[j],t = mul(a[j + h / 2],w);
		a[j] = inc(u,t);
		a[j + h / 2] = inc(u,MOD - t);
		w = mul(w,wn);
	    }
	}
    }
    if(on == -1) {
	int InvL = fpow(len,MOD - 2);
	for(int i = 0 ; i < len ; ++i) a[i] = mul(a[i],InvL);
    }
}
vector<int> operator - (vector<int> a,vector<int> b) {
    int s = max(a.size(),b.size());
    a.resize(s);b.resize(s);
    vector<int> c;c.clear();
    for(int i = 0 ; i < s ; ++i) c.pb(inc(a[i],MOD - b[i]));
    return c;
}
vector<int> operator + (vector<int> a,vector<int> b) {
    int s = max(a.size(),b.size());
    a.resize(s);b.resize(s);
    vector<int> c;c.clear();
    for(int i = 0 ; i < s ; ++i) c.pb(inc(a[i],b[i]));
    return c;
}
vector<int> operator * (vector<int> a,vector<int> b) {
    int t = a.size() + b.size() - 2,T = 1;
    while(T <= t) T <<= 1;
    vector<int> c;c.clear();
    NTT(a,T,1);NTT(b,T,1);
    for(int i = 0 ; i < T ; ++i) c.pb(mul(a[i],b[i]));
    NTT(c,T,-1);
    if(T > M + 1) c.resize(M + 1),T = M + 1;
    for(int i = T - 1 ; i > 0 ; --i) {
	if(!c[i]) c.pop_back();
	else break;
    }
    return c;
}
void dfs1(int u) {
    dep[u] = dep[fa[u]] + 1;
    siz[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
	int v = E[i].to;
	if(v != fa[u]) {
	    fa[v] = u;
	    dfs1(v);
	    siz[u] += siz[v];
	    if(siz[v] > siz[son[u]]) son[u] = v;
	}
    }
}
void dfs2(int u) {
    dfn[u] = ++tot;Line[tot] = u;
    ++cnt;
    if(!top[u]) top[u] = u;
    if(son[u]) {
	top[son[u]] = top[u];
	dfs2(son[u]);
    }
    else {
	lsiz[top[u]] = cnt;
	cnt = 0;
    }
    for(int i = head[u] ; i ; i = E[i].next) {
	int v = E[i].to;
	if(v != son[u] && v != fa[u]) dfs2(v);
    }
}
void Init() {
    W[0] = 1;W[1] = fpow(3,(MOD - 1) / L);
    for(int i = 2 ; i < L ; ++i) W[i] = mul(W[i - 1],W[1]);
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {
	read(B[i]);
	f[i][0].pb(1);
	f[i][1].pb(0),f[i][1].pb(B[i]);
    }
    int u,v;
    for(int i = 1 ; i < N ; ++i) {
	read(u);read(v);add(u,v);add(v,u);
    }
    dfs1(1);
    dfs2(1);
}
res_node DC(int l,int r) {
    if(l == r) {
	int u = Line[l];
	return (res_node){f[u][0],zero,zero,f[u][1]};
    }
    int mid = (l + r) >> 1;
    res_node wl = DC(l,mid),wr = DC(mid + 1,r);
    return (res_node){
	(wl.f00 + wl.f01) * (wr.f10 + wr.f00) - wl.f01 * wr.f10,
	(wl.f00 + wl.f01) * (wr.f11 + wr.f01) - wl.f01 * wr.f11,
	(wl.f10 + wl.f11) * (wr.f10 + wr.f00) - wl.f11 * wr.f10,
	(wl.f10 + wl.f11) * (wr.f01 + wr.f11) - wl.f11 * wr.f11,
    };
}
vector<int> mul(vector<int> *g,int l,int r) {
    if(l == r) return g[l];
    int mid = (l + r) >> 1;
    return mul(g,l,mid) * mul(g,mid + 1,r);
}
void Solve() {
    res_node t;
    for(int i = N ; i >= 1 ; --i) {
	int u = Line[i];
	if(top[u] == u) {
	    for(int j = dfn[u] ; j <= dfn[u] + lsiz[u] - 1 ; ++j) {
		int tot = 0;
		int c = Line[j];
		for(int k = head[c] ; k ; k = E[k].next) {
		    int v = E[k].to;
		    if(v != fa[c] && v != son[c]) g[0][++tot] = f[v][0] + f[v][1],g[1][tot] = f[v][0];
		}
		if(!tot) continue;
		f[c][0] = mul(g[0],1,tot);
		f[c][1] = f[c][1] * mul(g[1],1,tot);
	    }
	    t = DC(dfn[u],dfn[u] + lsiz[u] - 1);
	    f[u][0] = t.f00 + t.f01;
	    f[u][1] = t.f10 + t.f11;
	}
    }
    f[1][0].resize(M + 1);f[1][1].resize(M + 1);
    out(inc(f[1][0][M],f[1][1][M]));enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
    return 0;
}