【LOJ】#2230. 「BJOI2014」大融合

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].rev
    void 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 ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}