CC Prime Distance On Tree (樹的點分治 + FFT)
轉載請註明出處,謝謝http://blog.csdn.net/ACM_cxlove?viewmode=contents by---cxlove
題目:給出一棵樹,問有選出一個二元組(u , v),滿足u 到 v的路徑長度為素數的概率為多少。所有邊長度為1。
http://www.codechef.com/AUG13/problems/PRIMEDST
自從做了男人八題之後,覺得這種型別有點爛大街了。。。肯定是個點分治。
對於子樹預處理出所有的點到根的路徑長度之後,便是經典的處理選出兩個點,路徑之和長度為素數的種數。
不同於之前的題,可以二分,或者two points,甚至資料結構維護。
這是多項式乘法,一次卷積運算之後,然後 暴力列舉素數,統計總數。
鑑於最近FFT也有點爛大街的感覺,所以就很容易想到了。。
幸運的1A,感覺資料是不是略水。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
#include <queue>
#define mem(a , b) memset(a , b , sizeof(a))
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define lson step << 1
#define rson step << 1 | 1
using namespace std;
typedef long long LL;
const int N = 50005;
//FFT copy from kuangbin
const double pi = acos (-1.0);
// Complex z = a + b * i
struct Complex {
double a, b;
Complex(double _a=0.0,double _b=0.0):a(_a),b(_b){}
Complex operator + (const Complex &c) const {
return Complex(a + c.a , b + c.b);
}
Complex operator - (const Complex &c) const {
return Complex(a - c.a , b - c.b);
}
Complex operator * (const Complex &c) const {
return Complex(a * c.a - b * c.b , a * c.b + b * c.a);
}
};
//len = 2 ^ k
void change (Complex y[] , int len) {
for (int i = 1 , j = len / 2 ; i < len -1 ; i ++) {
if (i < j) swap(y[i] , y[j]);
int k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
// FFT
// len = 2 ^ k
// on = 1 DFT on = -1 IDFT
void FFT (Complex y[], int len , int on) {
change (y , len);
for (int h = 2 ; h <= len ; h <<= 1) {
Complex wn(cos (-on * 2 * pi / h), sin (-on * 2 * pi / h));
for (int j = 0 ; j < len ; j += h) {
Complex w(1 , 0);
for (int k = j ; k < j + h / 2 ; k ++) {
Complex u = y[k];
Complex t = w * y [k + h / 2];
y[k] = u + t;
y[k + h / 2] = u - t;
w = w * wn;
}
}
}
if (on == -1) {
for (int i = 0 ; i < len ; i ++) {
y[i].a /= len;
}
}
}
struct Edge {
int v , next;
}e[N << 1];
int n , tot , start[N];
int prime[N] , flag[N] , primecnt;
int del[N] , size[N];
LL ans = 0;
void Init (int n) {
mem (start , -1);
tot = 0;
primecnt = 0;
for (int i = 2 ; i < n ; i ++) {
if (flag[i]) continue;
prime[primecnt ++] = i;
for (int j = 2 ; j * i < n ; j ++)
flag[i * j] = 1;
}
}
void _add (int u , int v) {
e[tot].v = v; e[tot].next = start[u];
start[u] = tot ++;
}
void add (int u , int v) {
_add (u , v);
_add (v , u);
}
void cal (int u , int pre) {
size[u] = 1;
for (int i = start[u] ; i != -1 ; i = e[i].next) {
int v = e[i].v;
if (v == pre || del[v]) continue;
cal (v , u);
size[u] += size[v];
}
}
int newroot , maxsize , totalsize;
void dfs (int u , int pre) {
int mx = 0 , sum = 1;
for (int i = start[u] ; i != -1 ; i = e[i].next) {
int v = e[i].v;
if (v == pre || del[v]) continue;
dfs (v , u);
mx = max (mx , size[v]);
}
mx = max (mx , totalsize - size[u]);
if (mx < maxsize) maxsize = mx , newroot = u;
}
int search (int r) {
cal (r , -1);
totalsize = size[r];
maxsize = 1 << 30;newroot = -1;
dfs (r , -1);
return newroot;
}
int dist[N] , idx , cnt[N];
vector <int> sub[N] , all;
Complex x1[N << 2];
LL sum [N << 2];
LL fuck (vector<int> &v) {
if (v.size() == 0) return 0;
int len = v[v.size() - 1] + 1;
for (int i = 0 ; i < len ; i ++) {
cnt[i] = 0;
}
for (int i = 0 ; i < v.size() ; i ++) {
cnt[v[i]] ++ ;
}
int l = 1;
while (l < len * 2) l <<= 1;
for (int i = 0 ; i < len ; i ++) {
x1[i] = Complex (cnt[i] , 0);
}
for (int i = len ; i < l ; i ++) {
x1[i] = Complex (0 , 0);
}
FFT (x1 , l , 1);
for (int i = 0 ; i < l ; i ++) {
x1[i] = x1[i] * x1[i];
}
FFT (x1 , l , -1);
l = 2 * v[v.size() - 1];
LL ans = 0LL;
for (int i = 0 ; i <= l ; i ++) {
sum[i] = (LL)(x1[i].a + 0.5);
}
for (int i = 0 ; i < v.size() ; i ++)
sum[v[i] << 1] --;
for (int i = 0 ; i <= l ; i ++) {
sum[i] /= 2;
}
for (int i = 0 ; i < primecnt && prime[i] <= l ; i ++) {
ans += sum[prime[i]];
}
return ans;
}
void gao (int u , int pre) {
all.push_back (dist[u]);
sub[idx].push_back (dist[u]);
for (int i = start[u] ; i != -1 ; i = e[i].next) {
int v = e[i].v;
if (v == pre || del[v]) continue;
dist[v] = dist[u] + 1;
gao (v , u);
}
}
void solve (int root) {
root = search (root);
del[root] = 1;
if (totalsize == 1) return ;
idx = 0 ; all.clear();
for (int i = start[root] ; i != - 1 ; i = e[i].next) {
int v = e[i].v;
if (del[v]) continue;
sub[idx].clear();
dist[v] = 1;
gao (v , -1);
sort (sub[idx].begin() , sub[idx].end());
idx ++;
}
all.push_back(0);
sort (all.begin() , all.end());
ans += fuck (all);
for (int i = 0 ; i < idx ; i ++) {
ans -= fuck (sub[i]);
}
for (int i = start[root] ; i != -1 ; i = e[i].next) {
int v = e[i].v;
if (del[v]) continue;
solve (v);
}
}
int main () {
//freopen ("input.txt" , "r" , stdin);
scanf ("%d" , &n);
Init(n);
for (int i = 1 ; i < n ; i ++) {
int u , v;
scanf ("%d %d" , &u , &v);
add (u , v);
}
solve (1);
printf ("%.6f\n" , ans / (n * 0.5 * (n - 1)));
return 0;
}
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