Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 415 Accepted Submission(s): 172
/*
hdu 5398 动态树LCT
problem:
给你n个点,让你构成一棵树,边的权值为端点的最大公约数. 求树的最大边权和
solve:
枚举所有点,然后贪心的思想.如果n-1个点构成了一个最大的生成树,那么加入第n个点时我们应尽可能把它往
自己的约数上连(大->小).这样的话就在连接第二个约数的时候就会构成环,所以需要将 当前点-->当前约数点的链上
断开最小的一条边.
所以可以枚举 点和其约数,然后利用LCT维护链上面的最小边. 就这个想了很久,不知道应该吧边权保存在哪个端点
比较好,总感觉会有矛盾 = =||. 结果发现完全可以弄个点来替代边,将点上的值初始化无限大就行了T_T,好2...
hhh-2016-08-21 16:43:04
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson ch[0]
#define rson ch[1]
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 100100;
const int INF = 0x3f3f3f3f;
struct Node* null;
struct Node
{
Node* ch[2] ;
Node* fa;
int Min;
Node* Minnode;
int Size,lpos,rpos ;
int rev;
void newnode(int v)
{
Min = v;
Size = 1 ;
Minnode = this;
lpos = rpos = 0;
fa = ch[0] = ch[1] = null ;
rev = 0;
}
void ini()
{
Min = INF;
Size = 1 ;
Minnode = this;
lpos = rpos = 0;
fa = ch[0] = ch[1] = null ;
rev = 0;
}
void update_rev()
{
if(this == null)
return ;
swap(ch[0],ch[1]);
rev ^= 1;
}
void push_up ()
{
Minnode = this;
if(ch[0] && ch[0]->Minnode->Min < Minnode->Min)
Minnode = ch[0]->Minnode;
if(ch[1] && ch[1]->Minnode->Min < Minnode->Min)
Minnode = ch[1]->Minnode;
}
void push_down()
{
if(rev)
{
ch[0]->update_rev();
ch[1]->update_rev();
rev = 0;
}
}
void link_child ( Node* o , int d )
{
ch[d] = o ;
o->fa = this ;
}
int isroot()
{
return fa == null || this != fa->ch[0] && this != fa->ch[1] ;
}
void sign_down ()
{
if ( !isroot () ) fa->sign_down () ;
push_down () ;
}
void Rotate ( int d )
{
Node* p = fa ;
Node* gp = fa->fa ;
p->link_child ( ch[d] , !d ) ;
if ( !p->isroot () )
{
if ( gp->ch[0] == p ) gp->link_child ( this , 0 ) ;
else gp->link_child ( this , 1 ) ;
}
else fa = gp ;
link_child ( p , d ) ;
p->push_up () ;
}
void splay ()
{
sign_down () ;
while ( !isroot () )
{
if ( fa->isroot () )
{
this == fa->ch[0] ? Rotate ( 1 ) : Rotate ( 0 ) ;
}
else
{
if ( fa == fa->fa->ch[0] )
{
this == fa->ch[0] ? fa->Rotate ( 1 ) : Rotate ( 0 ) ;
Rotate ( 1 ) ;
}
else
{
this == fa->ch[1] ? fa->Rotate ( 0 ) : Rotate ( 1 ) ;
Rotate ( 0 ) ;
}
}
}
push_up () ;
}
void access()
{
Node* o = this ;
Node* x = null ;
while ( o != null )
{
o->splay () ;
o->link_child ( x , 1 ) ;
o->push_up () ;
x = o ;
o = o->fa ;
}
splay () ;
}
void make_root()
{
access();
update_rev();
}
void cut()
{
access();
ch[0]->fa = null;
ch[0] = null;
push_up();
}
Node* find_root ()
{
access () ;
Node* o = this ;
while ( o->ch[0] != null )
{
o->push_down () ;
o = o->ch[0] ;
}
return o ;
}
void cut(Node* to)
{
to->make_root();
cut();
}
void link(Node* to)
{
to->make_root();
to->fa = this;
}
Node* query(Node* to)
{
to->make_root();
access();
return Minnode;
}
};
Node memory_pool[maxn*2];
Node* now;
Node* node[maxn];
void Clear()
{
now = memory_pool;
now->newnode(INF);
null = now ++;
null->Size = 0;
}
int dp[maxn];
vector<int > vec[maxn];
void init()
{
Clear();
for(int i = 100000; i >= 1; i--)
{
now->newnode(INF);
node[i] = now++;
for(int j = i+i; j <= 100000; j+=i)
{
vec[j].push_back(i);
}
}
int tans = 0;
dp[1] = 0;
now->newnode(INF),node[1] = now++;
for(int i =2; i <= 100000; i++)
{
// cout << "number:" << i<<endl;
now->newnode(vec[i][0]);
Node *tp = now++;
tp->link(node[i]);
tp->link(node[vec[i][0]]);
tp->lpos = i,tp->rpos = vec[i][0];
tans += vec[i][0];
// cout << tp->lpos <<" " <<tp->rpos <<" " <<tans <<endl;
for(int j = 1;j < vec[i].size();j++)
{
int to = vec[i][j];
// cout <<"to "<<to <<endl;
Node* tMin = node[to]->query(node[i]);
// cout <<"Min:" << tMin->lpos <<" " <<tMin->rpos <<" "<<tMin->Min <<endl;
if(tMin->Min < to)
{
tans -= tMin->Min;
tMin->cut(node[tMin->lpos]);
// cout <<"cut1 ";
tMin->cut(node[tMin->rpos]);
// cout << "cut2" <<endl;
tMin->ini();
tMin->Min = to,tMin->lpos = i,tMin->rpos = to;
// cout << i <<" " <<to<<endl;
tMin->link(node[i]);
// cout <<"link1 ";
tMin->link(node[to]);
// cout << "link2" <<endl;
tans += to;
}
}
dp[i] = tans ;
}
}
int main()
{
// freopen("in.txt","r",stdin);
init();
int n;
while( scanf("%d",&n) !=EOF)
{
printf("%d\n",dp[n]);
}
return 0;
}
原文:http://www.cnblogs.com/Przz/p/5812330.html