题解

我现在真是太特么老年了

一写数据结构就颓废，难受

这题就是用lct维护子树
？？？lct怎么维护子树

这样想，我们给每个点记录虚边所在的子树大小，只发生在Access和link的时候
这样的话我们在这两个操作的时候顺带维护一下就好了

Access的时候加上新的虚儿子，减掉变成实边的那个儿子
link直接加上虚儿子的大小即可

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
int N,M;
namespace lct {struct node {int lc,rc,fa,sum,siz;bool rev;}tr[MAXN];
#define lc(u) tr[u].lc
#define rc(u) tr[u].rc
#define fa(u) tr[u].fa
#define sum(u) tr[u].sum
#define siz(u) tr[u].siz
#define rev(u) tr[u].revvoid Init() {for(int i = 1 ; i <= N ; ++i) sum(i) = siz(i) = 1;}void reverse(int u) {swap(lc(u),rc(u));rev(u) ^= 1;}void pushdown(int u) {if(rev(u)) {reverse(lc(u));reverse(rc(u));rev(u) = 0;}}void update(int u) {sum(u) = siz(u);sum(u) += sum(lc(u)) + sum(rc(u));}bool isRoot(int u) {if(!fa(u)) return true;else return rc(fa(u)) != u && lc(fa(u)) != u;} bool which(int u) {return u == rc(fa(u));}void rotate(int u) {int v = fa(u);if(!isRoot(v)) { (v == lc(fa(v)) ? lc(fa(v)) : rc(fa(v))) = u;}fa(u) = fa(v);fa(v) = u;if(u == lc(v)) {fa(rc(u)) = v;lc(v) = rc(u);rc(u) = v;}else {fa(lc(u)) = v;rc(v) = lc(u);lc(u) = v;}update(v);}void Splay(int u) {static int que[MAXN],qr;qr = 0;int x;for(x = u ; !isRoot(x) ; x = fa(x)) que[++qr] = x;que[++qr] = x;for(int i = qr ; i >= 1 ; --i) pushdown(que[i]);while(!isRoot(u)) {if(!isRoot(fa(u))) {if(which(fa(u)) == which(u)) rotate(fa(u));else rotate(u);}rotate(u);}update(u);}void Access(int u) {int x;for(x = 0 ; u ; x = u , u = fa(u)) {Splay(u);siz(u) += sum(rc(u));siz(u) -= sum(x);rc(u) = x;update(u);}}void Makeroot(int u) {Access(u);Splay(u);reverse(u);}void Link(int u,int v) {Makeroot(u);Makeroot(v);Splay(v);fa(v) = u;siz(u) += sum(v);Splay(u);}int64 Query(int u,int v) {Makeroot(u);Access(v);Splay(u);update(v);return 1LL * (sum(u) - sum(v)) * sum(v);}
}
void Solve() {char op[5];int x,y;read(N);read(M);lct::Init();for(int i = 1 ; i <= M ; ++i) {scanf("%s",op + 1);read(x);read(y);if(op[1] == 'A') {lct::Link(x,y);}else {out(lct::Query(x,y));enter;}}
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}