「bzoj3091」城市旅行 - LCT | Bill Yang's Blog

「bzoj3091」城市旅行 - LCT

题目大意


题目分析

将期望$E(x,y)$的式子列出来:
若将$(x,y)$的路径看作一个序列,$num$为路径上的结点数,则有:

这样我们就可以使用LCT提取路径,维护如下属性:

  • $val$:点权$a_i$
  • $sum$:点权和$\sum a_i$
  • $lazy$:懒标记
  • $small(x)$:对于$x$子树结点结点$i$,$\sum rank(i)*a_i$,$rank$表示结点在$x$子树中的排名(比它小的$+1$)
  • $big(x)$:对于$x$子树结点结点$i$,$\sum Mrank(i)*a_i$,$Mrank$表示结点在$x$子树中的从大到小排名(比它大的$+1$)
  • $ans(x)$:$\sum rank(i)Mrank(i)a_i$

然后考虑一下标记合并、标记上下传即可。

注意标记上传的时候子树的$rank$和$Mrank$一直在改变,需要维护。
注意好像要爆long long


代码

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#include<algorithm>
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
using namespace std;
typedef int int_;
#define int unsigned long long
inline const int Get_Int() {
int num=0,bj=1;
char x=getchar();
while(x<'0'||x>'9') {
if(x=='-')bj=-1;
x=getchar();
}
while(x>='0'&&x<='9') {
num=num*10+x-'0';
x=getchar();
}
return num*bj;
}
const int maxn=50005;
struct Tree {
int child[2],father,size;
bool rev;
int val,sum,lazy,small,big,ans;
};
struct Link_Cut_Tree {
Tree tree[maxn];
stack<int>S;
#define fa(x) tree[x].father
#define ls(x) tree[x].child[0]
#define rs(x) tree[x].child[1]
#define rev(x) tree[x].rev
#define val(x) tree[x].val
#define big(x) tree[x].big
#define sum(x) tree[x].sum
#define ans(x) tree[x].ans
#define lazy(x) tree[x].lazy
#define size(x) tree[x].size
#define small(x) tree[x].small
bool isroot(int index) {
return ls(fa(index))!=index&&rs(fa(index))!=index;
}
bool checkson(int index) {
return rs(fa(index))==index;
}
void modify(int index,int v) {
val(index)+=v;
lazy(index)+=v;
sum(index)+=size(index)*v;
small(index)+=size(index)*(size(index)+1)/2*v;
big(index)+=size(index)*(size(index)+1)/2*v;
ans(index)+=size(index)*(size(index)+1)*(size(index)+2)/6*v;
}
void rever(int index) {
rev(index)^=1;
swap(ls(index),rs(index));
swap(small(index),big(index));
}
void push_down(int index) {
if(rev(index)) {
if(ls(index))rever(ls(index));
if(rs(index))rever(rs(index));
rev(index)=0;
}
if(lazy(index)) {
if(ls(index))modify(ls(index),lazy(index));
if(rs(index))modify(rs(index),lazy(index));
lazy(index)=0;
}
}
void push_up(int index) {
int ls=ls(index),rs=rs(index);
size(index)=size(ls)+size(rs)+1;
sum(index)=sum(ls)+sum(rs)+val(index);
small(index)=small(ls)+small(rs)+(size(ls)+1)*sum(rs)+val(index)*(size(ls)+1);
big(index)=big(ls)+(size(rs)+1)*sum(ls)+big(rs)+val(index)*(size(rs)+1);
ans(index)=ans(ls)+(size(rs)+1)*small(ls)+ans(rs)+(size(ls)+1)*big(rs)+(size(ls)+1)*(size(rs)+1)*val(index);
}
void rotate(int index) {
int father=fa(index),grand=fa(father),side=checkson(index);
if(!isroot(father))tree[grand].child[checkson(father)]=index;
tree[father].child[side]=tree[index].child[side^1];
fa(tree[father].child[side])=father;
fa(father)=index;
tree[index].child[side^1]=father;
fa(index)=grand;
push_up(father);
push_up(index);
}
void splay(int index) {
S.push(index);
for(int i=index; !isroot(i); i=fa(i))S.push(fa(i));
while(!S.empty())push_down(S.top()),S.pop();
for(int father; !isroot(index); rotate(index)) {
father=fa(index);
if(!isroot(father))rotate(checkson(index)==checkson(father)?father:index);
}
}
void access(int index) {
for(int son=0; index; son=index,index=fa(index)) {
splay(index);
rs(index)=son;
push_up(index);
}
}
void reverse(int index) {
access(index);
splay(index);
rev(index)^=1;
swap(ls(index),rs(index));
}
void link(int x,int y) {
reverse(x);
fa(x)=y;
}
void split(int x,int y) {
reverse(x);
access(y);
splay(y);
}
void cut(int x,int y) {
split(x,y);
if(size(y)>2)return;
ls(y)=fa(x)=0;
}
int get_root(int index) {
access(index);
splay(index);
int u=index;
while(ls(u))push_down(u),u=ls(u);
push_down(u);
return u;
}
void modify(int x,int y,int v) {
split(x,y);
modify(y,v);
}
pair<int,int> query(int x,int y) {
split(x,y);
// debug(y);
int u=ans(y),d=size(y)*(size(y)+1)/2;
int gcd=__gcd(u,d);
u/=gcd,d/=gcd;
return make_pair(u,d);
}
void debug(int x) {
if(ls(x))debug(ls(x));
cout<<x<<" "<<fa(x)<<" "<<ls(x)<<" "<<rs(x)<<" "<<small(x)<<" "<<big(x)<<" "<<ans(x)<<endl;
if(rs(x))debug(rs(x));
}
} lct;
int n,q;
int_ main() {
n=Get_Int();
q=Get_Int();
for(int i=1; i<=n; i++)lct.tree[i].val=Get_Int();
for(int i=1; i<n; i++)lct.link(Get_Int(),Get_Int());
while(q--) {
int opt=Get_Int(),x=Get_Int(),y=Get_Int();
if(opt==1)lct.cut(x,y);
else if(opt==2) {
if(lct.get_root(x)==lct.get_root(y))continue;
lct.link(x,y);
} else if(opt==3) {
int v=Get_Int();
if(lct.get_root(x)!=lct.get_root(y))continue;
lct.modify(x,y,v);
} else {
if(lct.get_root(x)!=lct.get_root(y)) {
puts("-1");
continue;
}
pair<int,int>tmp=lct.query(x,y);
printf("%llu/%llu\n",tmp.first,tmp.second);
}
}
return 0;
}
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