【题解】【雅礼集训 2017 Day5】远行 LOJ 6038 LCT

Prelude

快要THUWC了，练一练板子。

传送到LOJ：o(TヘTo)

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Solution

首先有一条定理。

到树中任意一点的最远点一定是直径的两个端点之一。

我也不会证反正大家都在用，似乎可以用反证法搞一搞？

然后就是LCT和并查集随便做了。

对于每个连通块，只需要保存这个连通块的直径的两个端点就可以了。

然后合并两个连通块的时候更新一下。

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Code

#include <cstring>
#include <algorithm>
#include <cstdio>
#include <cctype>

using namespace std;
const int N = 300010;
int _w;

int read() {
	int x = 0, ch;
	while( isspace(ch = getchar()) );
	do x = x * 10 + ch - '0';
	while( isdigit(ch = getchar()) );
	return x;
}

int type, n, q, lastans;

namespace LCT {
	struct Node {
		int sz;
		Node *ch[2], *pa, *pathpa;
		bool rev;
		Node() {
			ch[0] = ch[1] = pa = pathpa = NULL;
			sz = 1, rev = 0;
		}
		int relation() {
			return this == pa->ch[0] ? 0 : 1;
		}
		Node *pushdown() {
			if( rev ) {
				rev = 0;
				swap( ch[0], ch[1] );
				if( ch[0] ) ch[0]->rev ^= 1;
				if( ch[1] ) ch[1]->rev ^= 1;
			}
			return this;
		}
		Node *maintain() {
			sz = 1;
			if( ch[0] ) sz += ch[0]->sz;
			if( ch[1] ) sz += ch[1]->sz;
			return this;
		}
		Node *rotate() {
			if( pa->pa ) pa->pa->pushdown();
			pa->pushdown(), pushdown();
			Node *old = pa;
			int x = relation();
			if( pa->pa ) pa->pa->ch[old->relation()] = this;
			pa = pa->pa;
			old->ch[x] = ch[x^1];
			if( ch[x^1] ) ch[x^1]->pa = old;
			ch[x^1] = old, old->pa = this;
			swap(old->pathpa, pathpa);
			return old->maintain(), maintain();
		}
		Node *splay() {
			while( pa ) {
				if( !pa->pa ) rotate();
				else {
					pa->pa->pushdown(), pa->pushdown();
					if( relation() == pa->relation() )
						pa->rotate(), rotate();
					else rotate(), rotate();
				}
			}
			return this;
		}
		Node *expose() {
			Node *rc = splay()->pushdown()->ch[1];
			if( rc ) {
				ch[1] = rc->pa = NULL;
				rc->pathpa = this;
				maintain();
			}
			return this;
		}
		bool splice() {
			if( !splay()->pathpa ) return false;
			pathpa->expose()->ch[1] = this;
			pa = pathpa, pathpa = NULL;
			pa->maintain();
			return true;
		}
		Node *access() {
			expose();
			while( splice() );
			return this;
		}
		Node *evert() {
			access()->rev ^= 1;
			return this;
		}
	};
	Node *rt[N];
	void init() {
		for( int i = 1; i <= n; ++i )
			rt[i] = new Node;
	}
	void link( int u, int v ) {
		rt[u]->evert()->pathpa = rt[v];
	}
	int query( int u, int v ) {
		rt[u]->evert();
		return rt[v]->access()->sz - 1;
	}
}

namespace DSU {
	int pa[N], du[N], dv[N];
	void init() {
		for( int i = 1; i <= n; ++i )
			pa[i] = du[i] = dv[i] = i;
	}
	int find( int u ) {
		return pa[u] == u ? u : pa[u] = find( pa[u] );
	}
	int uni( int u, int v ) {
		u = find(u), v = find(v);
		return pa[u] = v;
	}
}

namespace Solve {
	void init() {
		DSU::init();
		LCT::init();
	}
	void link( int u, int v ) {
		using DSU::du;
		using DSU::dv;
		using DSU::find;
		int u1 = du[find(u)], u2 = dv[find(u)];
		int v1 = du[find(v)], v2 = dv[find(v)];
		int w1 = LCT::query(u, u1) > LCT::query(u, u2) ? u1 : u2;
		int w2 = LCT::query(v, v1) > LCT::query(v, v2) ? v1 : v2;
		LCT::link(u, v);
		int rt = DSU::uni(u, v);
		int lenu = LCT::query(u1, u2);
		int lenv = LCT::query(v1, v2);
		int lenw = LCT::query(w1, w2);
		// printf( "w1 = %d, w2 = %d, lenw = %d\n", w1, w2, lenw );
		if( lenu >= lenv && lenu >= lenw )
			du[rt] = u1, dv[rt] = u2;
		else if( lenv >= lenu && lenv >= lenw )
			du[rt] = v1, dv[rt] = v2;
		else
			du[rt] = w1, dv[rt] = w2;
		// printf( "du = %d, dv = %d\n", du[rt], dv[rt] );
	}
	int query( int u ) {
		using DSU::du;
		using DSU::dv;
		using DSU::find;
		int u1 = du[find(u)], u2 = dv[find(u)];
		return max( LCT::query(u, u1), LCT::query(u, u2) );
	}
}

int main() {
	type = read(), n = read(), q = read();
	Solve::init();
	while( q-- ) {
		if( read() == 1 ) {
			int u = read(), v = read();
			u ^= type * lastans;
			v ^= type * lastans;
			Solve::link(u, v);
		} else {
			int u = read();
			u ^= type * lastans;
			printf( "%d\n", lastans = Solve::query(u) );
		}
	}
	return 0;
}