hdu 5398 GCD Tree 2015多校联合训练赛#9 LCT，动态生成树

GCD Tree

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 122    Accepted Submission(s): 56

Problem Description

Teacher Mai has a graph with n vertices numbered from 1 to n. For every edge(u,v), the weight is gcd(u,v). (gcd(u,v) means the greatest common divisor of number u and v).

You need to find a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is maximized. Print the total weight of these edges.

 

Input

There are multiple test cases(about 105).

For each test case, there is only one line contains one number n(1≤n≤105).

 

Output

For each test case, print the answer.

 

Sample Input

12345

 

Sample Output

01245

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

 

解题报告已经如此清晰了。我就不多说了。

讲讲LCT吧。这里每个点作为LCT的一个结点，每条边作为LCT的一个结点。于是删除边的时候只要找到对应的点删除即可。

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>using namespace std;#define maxn 2000007#define inf  1000000000#define ll long longstruct Node{    Node *fa,*ch[2];    bool rev,root;    int val,minv;};Node pool[maxn];Node *nil,*tree[maxn];int cnt = 0;void init(){    cnt = 1;    nil = tree[0] = pool;    nil->ch[0] = nil->ch[1] = nil;    nil->val = inf;    nil->minv = inf;}Node *newnode(int val,Node *f){    pool[cnt].fa = f;    pool[cnt].ch[0]=pool[cnt].ch[1]=nil;    pool[cnt].rev = false;    pool[cnt].root = true;    pool[cnt].val = val;    pool[cnt].minv = val;    return &pool[cnt++];}//左右子树反转******真正把结点变为根void update_rev(Node *x){    if(x == nil) return ;    x->rev = !x->rev;    swap(x->ch[0],x->ch[1]);}//splay向上更新信息******void update(Node *x){    if(x == nil) return ;    x->minv = x->val;    if(x->ch[0] != nil)        x->minv = min(x->minv,x->ch[0]->minv);    if(x->ch[1] != nil)        x->minv = min(x->minv,x->ch[1]->minv);}//splay下推信息******void pushdown(Node *x){    if(x->rev != false){        update_rev(x->ch[0]);        update_rev(x->ch[1]);        x->rev = false;    }}//splay在root-->x的路径下推信息******void push(Node *x){    if(!x->root) push(x->fa);    pushdown(x);}//将结点x旋转至splay中父亲的位置******void rotate(Node *x){    Node *f = x->fa, *ff = f->fa;    int t = (f->ch[1] == x);    if(f->root)        x->root = true, f->root = false;    else ff->ch[ff->ch[1] == f] = x;    x->fa = ff;    f->ch[t] = x->ch[t^1];    x->ch[t^1]->fa = f;    x->ch[t^1] = f;    f->fa = x;    update(f);}//将结点x旋转至x所在splay的根位置******void splay(Node *x){    push(x);    Node *f, *ff;    while(!x->root){        f = x->fa,ff = f->fa;        if(!f->root)            if((ff->ch[1]==f)&&(f->ch[1] == x)) rotate(f);            else rotate(x);        rotate(x);    }    update(x);}//将x到树根的路径并成一条path******Node *access(Node *x){    Node *y = nil,*z;    while(x != nil){        splay(x);        x->ch[1]->root = true;        (x->ch[1] = y)->root = false;        update(x);        y = x;        x = x->fa;    }    return y;}//将结点x变成树根******void be_root(Node *x){    access(x);    splay(x);    update_rev(x);}ll ans[maxn];//将x连接到结点f上******void link(Node *x, Node *f){    be_root(x);    x->fa = f;    access(x);}//将x,y分离******void cut(Node *x,Node *y){    be_root(x);    access(x);    splay(y);    y->fa = nil;}Node * find_fa(Node *root){    if(root->ch[0] == nil) return root;    return find_fa(root->ch[0]);}Node * find_min(Node * root,int val){    if(root->val == val) return root;    if(root->ch[0]->minv == val)        return find_min(root->ch[0],val);    return find_min(root->ch[1],val);}//处理u到v是否连边int work(int u,int v){    Node *x,*y,*z;    be_root(tree[u]);    y = access(tree[v]);    x = find_fa(y);    int res = 0;    if(x == tree[u]){        if(y->minv >= v) return 0;        y = find_min(y,y->minv);        cut(y,tree[u]);        cut(y,tree[v]);        res -= y->val;    }    y = newnode(v,nil);    link(tree[u],y);    link(tree[v],y);    return res+v;}int main(){    memset(ans,0,sizeof(ans));    init();    int u,v;    for(int i = 1;i <= 100000;i++){        ans[i] = ans[i-1];        tree[i] = newnode(inf,nil);        for(int j = 1;j * j <= i && j < i; j++){            if(i%j != 0) continue;            ans[i]+=work(i,j);            if(i/j != j) ans[i]+=work(i,i/j);        }    }//    freopen("in.txt","r",stdin);//    freopen("outx.txt","w",stdout);    int n;    while(scanf("%d",&n)!=EOF)        printf("%I64d\n",ans[n]);    return 0;}