bzoj 3091: 城市旅行 LCT

题目:

http://www.lydsy.com/JudgeOnline/problem.php?id=3091

题解:

首先前三个操作就是裸的LCT模板
只考虑第四个操作.
要求我们计算期望,所以我们考虑计算出所有情况的和然后在除以情况的数目.
这样我们就找到分子分母了.
我们很容易发现分母即为(frac{n*(n+1)}{2})
对应到我们的Splay树上即(frac{siz*(siz+1)}{2})

所以我们现在考虑维护分子:
对于首先我们考虑在一个长为n的序列上统计这些东西
我们知道总和即为每一项乘以这一项出现的次数(又在废话)
出现的次数又是多少呢?

[ egin{matrix} a_1 & a_2 & a_3 & ... & a_i &... & a_n \ 1*n & 2*(n-1) & 3*(n-2) & ...&i*(n-i+1) & ...& n*1 \ end{matrix} ]

所以其实对于每一个元素,出现的次数都是(( ext{左边的}siz+1)*( ext{右边的}siz+1))
那么我们考虑合并:
假设这个区间作为合并的左区间,我们设(w = ext{右区间的}siz+1)(即这个区间合并后右侧新出现的节点数)
那么按照刚才的思路,所有的数字的后一项都会同时增大即变为:

[ egin{matrix} a_1 & a_2 & a_3 & ... & a_i &... & a_n \ 1*(n+w) & 2*(n-1+w) & 3*(n-2+w) & ...&i*(n-i+1+w) & ...& n*(1+w) \ end{matrix} ]

于是我们发现实际上这段区间的贡献增加了:

[ egin{matrix} a_1 & a_2 & a_3 & ... & a_i &... & a_n \ 1*w & 2*w & 3*w & ...& i*w & ...& n*w \ end{matrix} ]

所以我们记录一个和表示(1*a_1 + 2*a_2 + 3*a_3 + ... + n*a_n)即可
利用这个我们就可以维护分子了。啥?? 怎么维护 ??
(val = ch[0]->val + ch[1]->val + ch[0]->lsum*(ch[1]->siz + 1) + ch[1]->rsum*(ch[0]->siz + 1) + w*(ch[0]->siz + 1)*(ch[1]->siz + 1);)
其中(w)为节点本身的权,(lsum = sum_{i=1}^{n}a_i*i),(rsum = sum_{i=1}^{n}a_i*(n-i+1))
至于维护(lsum)(rsum)的过程.
我们有

[1*n + 2*(n-1) + 3*(n-2) + ... + n*1 = frac{n*(n+1)*(n+2)}{6} ]

据说是小学数学难度.
夭折啊 !我想不出来 !

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef unsigned long long ll;
inline void read(ll &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
const ll maxn = 50010;
struct Node{
    Node *ch[2],*fa;
    ll w,lsum,rsum,lazy;
    ll val,siz,tag,sum;
    void update();
    void pushdown();
    void rev();
    void inc(ll x);
}*null;
void Node::rev(){
    if(this == null) return;
    swap(lsum,rsum);swap(ch[0],ch[1]);
    tag ^= 1;
}
void Node::inc(ll x){
    if(this == null) return;
    w += x;sum += x*siz;
    lsum += x*siz*(siz+1)/2;
    rsum += x*siz*(siz+1)/2;
    val += x*siz*(siz+1)*(siz+2)/6;
    lazy += x;
}
void Node::pushdown(){
    if(this == null) return;
    if(lazy){
        if(ch[0] != null) ch[0]->inc(lazy);
        if(ch[1] != null) ch[1]->inc(lazy);
        lazy = 0;
    }
    if(tag){
        if(ch[0] != null) ch[0]->rev();
        if(ch[1] != null) ch[1]->rev();
        tag = 0;
    }
}
void Node::update(){
    if(this == null) return;
    siz = ch[0]->siz + ch[1]->siz + 1;
    sum = ch[0]->sum + ch[1]->sum + w;
    lsum = ch[0]->lsum + w*(ch[0]->siz + 1) + ch[1]->lsum + ch[1]->sum*(ch[0]->siz + 1);
    rsum = ch[1]->rsum + w*(ch[1]->siz + 1) + ch[0]->rsum + ch[0]->sum*(ch[1]->siz + 1);
    val = ch[0]->val + ch[1]->val + ch[0]->lsum*(ch[1]->siz + 1) + ch[1]->rsum*(ch[0]->siz + 1) + w*(ch[0]->siz + 1)*(ch[1]->siz + 1);
}
Node mem[maxn],*it;
inline void init(){
    it = mem;null = it++;
    null->ch[0] = null->ch[1] = null->fa = null;
    null->w = null->lsum = null->rsum = null->sum = 
    null->val = null->siz = null->tag = 0;
}
inline Node* newNode(ll x){
    Node *p = it++;p->ch[0] = p->ch[1] = p->fa = null;
    p->w = p->val = p->lsum = p->rsum = p->sum = x;p->siz = 1;
    p->tag = p->lazy = 0;
    return p;
}
inline void rotate(Node *p,Node *x){
    ll k = p == x->ch[1];
    Node *y = p->ch[k^1],*z = x->fa;
    if(z->ch[0] == x) z->ch[0] = p;
    if(z->ch[1] == x) z->ch[1] = p;
    if(y != null) y->fa = x;
    p->fa = z;p->ch[k^1] = x;
    x->fa = p;x->ch[k] = y;
    x->update();p->update();
}
inline bool isRoot(Node *p){
    return (p == null) || (p->fa->ch[0] != p && p->fa->ch[1] != p);
}
inline void Splay(Node *p){
    p->pushdown();
    while(!isRoot(p)){
        Node *x = p->fa,*y = x->fa;
        y->pushdown();x->pushdown();p->pushdown();
        if(isRoot(x)) rotate(p,x);
        else if((p == x->ch[0])^(x == y->ch[0])) rotate(p,x),rotate(p,y);
        else rotate(x,y),rotate(p,x);
    }p->update();
}
inline Node* Access(Node *x){
    for(Node *y = null;x != null;y = x,x = x->fa)
        Splay(x),x->ch[1] = y,x->update();
    return x;
}
inline void makeRoot(Node *x){
    Access(x);Splay(x);x->rev();
}
inline void link(Node *x,Node *y){
    makeRoot(x);x->fa = y;
}
inline void cut(Node *x,Node *y){
    makeRoot(x);Access(y);Splay(y);
    if(y->ch[0] == x && x->ch[1] == null){
        y->ch[0] = y->ch[0]->fa = null;
        y->update();
    }
}
inline void inc(Node *x,Node *y,ll w){
    makeRoot(x);Access(y);Splay(y);
    y->inc(w);
}
inline ll gcd(const ll &a,const ll &b){return b == 0 ? a : gcd(b,a%b);}
inline ll query(Node *x,Node *y){
    makeRoot(x);Access(y);Splay(y);
    ll upside = y->val;
    ll dnside = y->siz*(y->siz + 1)/2;
    ll g = gcd(upside,dnside);
    printf("%llu/%llu
",upside/g,dnside/g);
}
inline Node* findRoot(Node *x){
    Access(x);Splay(x);
    while(x->ch[0] != null) x = x->ch[0];
    Splay(x);return x;
}
int main(){
    init();
    ll n,m;read(n);read(m);
    for(ll i=1,x;i<=n;++i){
        read(x);newNode(x);
    }
    ll u,v;
    for(ll i=1;i<n;++i){
        read(u);read(v);
        link(mem+u,mem+v);
    }
    ll op;
    while(m--){
        read(op);
        if(op == 1){
            read(u);read(v);
            if(u != v && findRoot(mem+u) == findRoot(mem+v)) cut(mem+u,mem+v);
        }else if(op == 2){
            read(u);read(v);
            if(findRoot(mem+u) != findRoot(mem+v)) link(mem+u,mem+v);
        }else if(op == 3){
            read(u);read(v);read(op);
            if(findRoot(mem+u) == findRoot(mem+v)) inc(mem+u,mem+v,op);
        }else if(op == 4){
            read(u);read(v);
            if(findRoot(mem+u) == findRoot(mem+v)) query(mem+u,mem+v);
            else puts("-1");
        }
    }
    getchar();getchar();
    return 0;
}
原文地址:https://www.cnblogs.com/Skyminer/p/6550818.html