BZOJ3451 Normal 期望、点分治、NTT
题目大意：给出一棵树，求对其进行随机点分治的复杂度期望
可以知道一个点的贡献就是其点分树上的深度，也就是这个点在点分树上的祖先数量+1。
根据期望的线性性，考虑一个点对\((x,y)\)在何时\(x\)能够是\(y\)的祖先，那么在\(x\)到\(y\)的路径上的所有点中\(x\)必须要是第一个被选中的点，否则要么点\(y\)变成点\(x\)的祖先，要么点\(x\)和点\(y\)会被分在不同的子树中。那么我们要求的就是\(\sum \frac{1}{dis_{x,y}}\)，其中\(dis_{x,y}\)表示\(x\)到\(y\)路径上的点的数量。直接使用\(NTT\)统计路径长度即可。
注意在某一个分治中心进行合并时必须按照深度从小到大合并，否则复杂度是不正确的。
#include<bits/stdc++.h>
#define INF 0x7fffffff
using namespace std;
inline int read(){
int a = 0;
char c = getchar();
while(!isdigit(c))
c = getchar();
while(isdigit(c)){
a = a * 10 + c - 48;
c = getchar();
}
return a;
}
const int MAXN = 1e5 + 10 , MOD = 998244353 , G = 3 , INV = 332748118;
struct Edge{
int end , upEd;
}Ed[MAXN << 1];
int head[MAXN] , ind[MAXN] , size[MAXN] , ans[MAXN];
int N , need , nSize , mSize , mInd , cntEd;
bool vis[MAXN];
long long inv_need;
vector < vector < int > > v;
inline void addEd(int a , int b){
Ed[++cntEd].end = b;
Ed[cntEd].upEd = head[a];
head[a] = cntEd;
}
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;
}
void NTT(vector < int > &v , int type){
for(int i = 0 ; i < need ; ++i)
if(i < ind[i])
swap(v[i] , v[ind[i]]);
for(int i = 1 ; i < need ; i <<= 1){
int wn = poww(type == 1 ? G : INV , (MOD - 1) / (i << 1));
for(int j = 0 ; j < need ; j += i << 1){
long long w = 1;
for(int k = 0 ; k < i ; ++k , w = w * wn % MOD){
int x = v[j + k] , y = v[i + j + k] * w % MOD;
v[j + k] = x + y >= MOD ? x + y - MOD : x + y;
v[i + j + k] = x < y ? x - y + MOD : x - y;
}
}
}
}
void getSize(int x){
vis[x] = 1;
++nSize;
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(!vis[Ed[i].end])
getSize(Ed[i].end);
vis[x] = 0;
}
void getRoot(int x){
vis[x] = 1;
int maxN = 0;
size[x] = 1;
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(!vis[Ed[i].end]){
getRoot(Ed[i].end);
maxN = max(maxN , size[Ed[i].end]);
size[x] += size[Ed[i].end];
}
maxN = max(maxN , nSize - size[x]);
if(maxN < mSize){
mSize = maxN;
mInd = x;
}
vis[x] = 0;
}
void calc(vector < int > &v , int x , int l){
while(v.size() <= l)
v.push_back(0);
++v[l];
vis[x] = 1;
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(!vis[Ed[i].end])
calc(v , Ed[i].end , l + 1);
vis[x] = 0;
}
bool cmp(vector < int > a , vector < int > b){
return a.size() < b.size();
}
void solve(int x){
nSize = 0;
mSize = INF;
getSize(x);
getRoot(x);
x = mInd;
vis[x] = 1;
v.clear();
vector < int > cur , ano;
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(!vis[Ed[i].end]){
cur.clear();
calc(cur , Ed[i].end , 1);
v.push_back(cur);
}
cur.clear();
sort(v.begin() , v.end());
need = 1;
for(int i = 0 ; i < v.size() ; ++i){
while(need < cur.size() + v[i].size())
need <<= 1;
inv_need = poww(need , MOD - 2);
for(int j = 1 ; j < need ; ++j)
ind[j] = (ind[j >> 1] >> 1) | (j & 1 ? need >> 1 : 0);
while(cur.size() < need)
cur.push_back(0);
while(v[i].size() < need)
v[i].push_back(0);
ano.clear();
NTT(cur , 1);
NTT(v[i] , 1);
for(int j = 0 ; j < need ; ++j){
ano.push_back(1ll * cur[j] * v[i][j] % MOD);
cur[j] = cur[j] + v[i][j] >= MOD ? cur[j] + v[i][j] - MOD : cur[j] + v[i][j];
}
NTT(ano , -1);
NTT(cur , -1);
for(int j = 0 ; j < need ; ++j){
ans[j] += ano[j] * inv_need % MOD;
cur[j] = cur[j] * inv_need % MOD;
}
while(cur.back() == 0)
cur.pop_back();
}
for(int i = 1 ; i < need ; ++i)
ans[i] += cur[i];
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(!vis[Ed[i].end])
solve(Ed[i].end);
}
int main(){
N = read();
for(int i = 1 ; i < N ; ++i){
int a = read() , b = read();
addEd(a , b);
addEd(b , a);
}
solve(1);
long double sum = 0;
for(int i = 1 ; i <= N ; ++i)
sum += ans[i] * 2.0 / (i + 1);
cout << fixed << setprecision(4) << sum + N;
return 0;
}
