HDU 5398 GCD Tree
来源：互联网 发布：阿娇 知乎 编辑：程序博客网 时间：2024/04/24 09:23
这题可以基本说是LCT的模板题目，几乎没什么多余的考虑。不像HDU 5333，那题除了用LCT维护最大生成树之外还有一些复杂的公式推算。
对于多组数据，从1枚举到n，然后加入他向因子连的所有边，使用LCT维护最大生成树即可。
因子连边不知道随机顺序地连边能不能过，但是可以证明的是，如果现在枚举到
时间复杂度
// whn6325689// Mr.Phoebe// http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,r#define root 1,1,ntemplate<class T>inline bool read(T &n){ T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true;}template <class T>inline void write(T n){ if(n < 0) { putchar('-'); n = -n; } int len = 0,data[20]; while(n) { data[len++] = n%10; n /= 10; } if(!len) data[len++] = 0; while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=100010;const int MAXE=2000005;struct Edge{ int to, from; Edge () {} Edge ( int to , int from) : to (to) , from (from) {}} ;struct Node* null ;struct Node{ Node* c[2] ; Node* f ; bool flip ; int minv , val ; int eidx , idx ; void newnode ( int v , int i ) { c[0] = c[1] = f = null ; minv = val = v ; eidx = idx = i ; flip = 0 ; } void rev () { if ( this == null ) return ; swap ( c[0] , c[1] ) ; flip ^= 1 ; } void up () { if ( this == null ) return ; if ( val <= c[0]->minv && val <= c[1]->minv ) { minv = val ; eidx = idx ; } else if ( c[0]->minv <= c[1]->minv && c[0]->minv <= val ) { minv = c[0]->minv ; eidx = c[0]->eidx ; } else { minv = c[1]->minv ; eidx = c[1]->eidx ; } } void down () { if ( this == null ) return ; if ( flip ) { c[0]->rev () ; c[1]->rev () ; flip = 0 ; } } bool is_root () { return f == null || f->c[0] != this && f->c[1] != this ; } void sign_down () { if ( !is_root () ) f->sign_down () ; down () ; } void setc ( Node* o , int d ) { c[d] = o ; o->f = this ; } void rot ( int d ) { Node* p = f ; Node* g = f->f ; p->setc ( c[d] , !d ) ; if ( !p->is_root () ) g->setc ( this , f == g->c[1] ) ; else f = g ; setc ( p , d ) ; p->up () ; } void splay () { sign_down () ; while ( !is_root () ) { if ( f->is_root () ) rot ( this == f->c[0] ) ; else { if ( f == f->f->c[0] ) { if ( this == f->c[0] ) f->rot ( 1 ) , rot ( 1 ) ; else rot ( 0 ) , rot ( 1 ) ; } else { if ( this == f->c[1] ) f->rot ( 0 ) , rot ( 0 ) ; else rot ( 1 ) , rot ( 0 ) ; } } } up () ; } void access () { Node* o = this ; for ( Node* x = null ; o != null ; x = o , o = o->f ) { o->splay () ; o->setc ( x , 1 ) ; o->up () ; } splay () ; } void make_root () { access () ; rev () ; } void link ( Node* o ) { make_root () ; f = o ; } void cut () { access () ; c[0] = c[0]->f = null ; up () ; } void cut ( Node* o ) { make_root () ; o->cut () ; } int get_min ( Node* o ) { make_root () ; o->access () ; return o->eidx ; }} ;Node pool[MAXN+MAXE];Node* cur;Node* node[MAXN];Node* edge[MAXE];Edge E[MAXE+MAXN];int U[MAXE],V[MAXE];int n,idx;ll ans;ll dp[MAXN];vector<int> g[MAXN];void init(int n){ idx=0,ans=0; cur=pool; cur->newnode(INF,-1); null=cur++; for(int i=1;i<=n;i++) { cur->newnode(INF,-1); node[i]=cur++; }}void addedge(int u,int v){ int eidx=node[u]->get_min(node[v]); if(E[eidx].from >= u) return; U[idx]=u; V[idx]=v; cur->newnode(u,idx); edge[idx]=cur; E[idx]=Edge(v,u); edge[eidx]->cut(node[U[eidx]]); edge[eidx]->cut(node[V[eidx]]); edge[idx]->link(node[u]); edge[idx]->link(node[v]); ans+=u-E[eidx].from; idx++;cur++;}void init(){ init(100000); for(int i=2;i<=100000;i++) { g[i].pb(1); for(int j=i+i;j<=100000;j+=i) g[j].pb(i); } int sz; for(int i=2;i<=100000;i++) { sz=g[i].size(); U[idx]=g[i][sz-1]; V[idx]=i; cur->newnode(g[i][sz-1],idx); edge[idx]=cur; E[idx]=Edge(i,g[i][sz-1]); edge[idx]->link(node[g[i][sz-1]]); edge[idx]->link(node[i]); ans+=g[i][sz-1]; idx++;cur++; for(int j=sz-2;j>=0;j--) addedge(g[i][j],i); dp[i]=ans; }}int main(){ init(); int m; while(~scanf("%d",&m)) printf("%lld\n",dp[m]); return 0;} 贴上杜教的标程，简直比我短到不知道哪里去了
杜教的写法似乎是枚举到
#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <sstream>#include <map>#include <set>#include <queue>#include <stack>#include <fstream>#include <numeric>#include <iomanip>#include <bitset>#include <list>#include <stdexcept>#include <functional>#include <utility>#include <ctime>#include <cassert>#include <complex>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define mp make_pair#define all(x) (x).begin(),(x).end()#define fi first#define se second#define SZ(x) ((int)(x).size())#define ACCU accumulate#define TWO(x) (1<<(x))#define TWOL(x) (1ll<<(x))#define clr(a) memset(a,0,sizeof(a))#define POSIN(x,y) (0<=(x)&&(x)<n&&0<=(y)&&(y)<m)#define PRINTC(x) cout<<"Case #"<<++__<<": "<<x<<endl #define POP(x) (__builtin_popcount(x))#define POPL(x) (__builtin_popcountll(x))typedef vector<int> VI;typedef vector<string> VS;typedef vector<double> VD;typedef long long ll;typedef long double LD;typedef pair<int,int> PII;typedef pair<ll,ll> PLL;typedef vector<ll> VL;typedef vector<PII> VPII;typedef complex<double> CD;const int inf=0x20202020;const ll mod=1000000007;const double eps=1e-9;const double pi=3.1415926535897932384626;const int DX[]={1,0,-1,0},DY[]={0,1,0,-1};ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}// headconst int N=101000;struct node { node *s[2],*f; int val,d; bool rev; bool isr() { return !f||(f->s[0]!=this && f->s[1]!=this);} bool dir() { return f->s[1]==this;} void setc(node *c,int d) { s[d]=c;if (c) c->f=this;} void push() { if (rev) { swap(s[0],s[1]); rep(i,0,2) if (s[i]) s[i]->rev^=1;} rev=0; } void upd() { val=d; rep(i,0,2) if (s[i]&&s[i]->val<val) val=s[i]->val; }}pool[N],*cur;stack<node*> sta;void rot(node *x) { node *p=x->f;bool d=x->dir(); if (!p->isr()) p->f->setc(x,p->dir()); else x->f=p->f; p->setc(x->s[!d],d);x->setc(p,!d); p->upd();}void splay(node *x) { node *q=x; while (1) { sta.push(q);if (q->isr()) break; q=q->f; } while (!sta.empty()) sta.top()->push(),sta.pop(); while (!x->isr()) { if (x->f->isr()) rot(x); else if (x->isr()==x->f->isr()) rot(x->f),rot(x); else rot(x),rot(x); } x->upd();}node *expose(node *x) { node *q=NULL; for (;x;x=x->f) splay(x),x->s[1]=q,(q=x)->upd(); return q;}void evert(node *x) { expose(x); splay(x); x->rev^=1; x->push();}void expose(node *x,node *y) { evert(x); expose(y); splay(x);}void link(node *x,node *y) { evert(x); evert(y); x->setc(y,1);}void cut(node *x,node *y) { expose(x,y); x->s[1]=y->f=NULL;}const int R=100000;int ans,ret[N],n;VI d[N];int main() { for (int i=2;i<=R;i++) for (int j=i+i;j<=R;j+=i) d[j].pb(i); for (int i=1;i<=R;i++) { pool[i].d=pool[i].val=i; } for (int i=2;i<=R;i++) { link(pool+i,pool+1); ans+=1; per(j,0,SZ(d[i])) { int v=d[i][j]; expose(pool+i,pool+v); int u=pool[i].val; if (u>=v) continue; ans=ans-u+v; splay(pool+u); if (pool[u].s[0]) pool[u].s[0]->f=NULL,pool[u].s[0]=NULL; else if (pool[u].s[1]) pool[u].s[1]->f=NULL,pool[u].s[1]=NULL; else assert(0); link(pool+i,pool+v); } ret[i]=ans; } while (scanf("%d",&n)!=EOF) { printf("%d\n",ret[n]); }} 0 0
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