【題解】【雅禮集訓 2017 Day5】遠行 LOJ 6038 LCT

Prelude

快要THUWC了，練一練板子。

傳送到LOJ：o(TヘTo)

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Solution

首先有一條定理。
到樹中任意一點的最遠點必定是直徑的兩個端點之一。
我也不會證反正你們都在用，彷佛能夠用反證法搞一搞？
而後就是LCT和並查集隨便作了。
對於每一個連通塊，只須要保存這個連通塊的直徑的兩個端點就能夠了。
而後合併兩個連通塊的時候更新一下。

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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;
}