bzoj3123: [Sdoi2013]森林

题面传送门

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复出的第一道题.. md就遇到坑了..

简单来说就是可持久化线段树+启发式合并啊..

感觉启发式合并好神奇好想学

每一次建边就暴力合并，每一个节点维护从根到它的权值线段树

按照题面的话最省空间的做法就是垃圾回收，但是实在是太慢了..

而且这题有坑，题面说的是多组数据其实只有一组 而且是$T>1$的一组..

然后看给了512MB就不需要垃圾回收，而且很多预处理都tm不用了呢！wqnmdsy

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int Maxn = 80010;
const int lg = 18;
struct node {
	int y, next;
}a[Maxn*2]; int first[Maxn], len;
void ins ( int x, int y ){
	len ++;
	a[len].y = y;
	a[len].next = first[x]; first[x] = len;
}
int n, m, t;
int sum[Maxn*lg*lg], lc[Maxn*lg*lg], rc[Maxn*lg*lg], tot;
int rt[Maxn];
int na[Maxn], b[Maxn], bl;
int fa[Maxn][lg], dep[Maxn];
int faa[Maxn], gs[Maxn];
int ff ( int x ){
	if ( faa[x] == x ) return x;
	return faa[x] = ff (faa[x]);
}
void merge ( int &now, int fnow, int l, int r, int x ){
	if ( !now ) now = ++tot;
	sum[now] = sum[fnow]+1;
	if ( l == r ) return;
	int mid = ( l + r ) >> 1;
	if ( x <= mid ) merge ( lc[now], lc[fnow], l, mid, x ), rc[now] = rc[fnow];
	else merge ( rc[now], rc[fnow], mid+1, r, x ), lc[now] = lc[fnow];
}
void dfs1 ( int x, int f ){
	rt[x] = 0;
	merge ( rt[x], rt[fa[x][0]], 1, bl, na[x] );
	for ( int i = 1; i <= 17; i ++ ) fa[x][i] = fa[fa[x][i-1]][i-1];
	for ( int k = first[x]; k; k = a[k].next ){
		int y = a[k].y;
		if ( y == f ) continue;
		fa[y][0] = x; dep[y] = dep[x]+1;
		dfs1 ( y, x );
	}
}
void change ( int &now, int l, int r, int x ){
	if ( !now ) now = ++tot;
	sum[now] ++;
	if ( l == r ) return;
	int mid = ( l + r ) >> 1;
	if ( x <= mid ) change ( lc[now], l, mid, x );
	else change ( rc[now], mid+1, r, x );
}
int getlca ( int x, int y ){
	if ( dep[x] < dep[y] ) swap ( x, y );
	for ( int i = 17; i >= 0; i -- ){
		if ( dep[fa[x][i]] >= dep[y] ){
			x = fa[x][i];
		}
	}
	if ( x == y ) return x;
	for ( int i = 17; i >= 0; i -- ){
		if ( fa[x][i] != fa[y][i] ){
			x = fa[x][i]; y = fa[y][i];
		}
	}
	return fa[x][0];
}
int getrank ( int now1, int now2, int now3, int p, int l, int r, int k ){
	if ( l == r ) return b[l];
	int mid = ( l + r ) >> 1;
	int ls = sum[lc[now1]]+sum[lc[now2]]-2*sum[lc[now3]];
	if ( na[p] <= mid && na[p] >= l ) ls --;
	if ( ls >= k ) return getrank ( lc[now1], lc[now2], lc[now3], p, l, mid, k );
	else return getrank ( rc[now1], rc[now2], rc[now3], p, mid+1, r, k-ls );
}
void read ( int &x ){
	char c = getchar ();
	for ( ; c > '9' || c < '0'; c = getchar () );
	x = 0;
	for ( ; c <= '9' && c >= '0'; c = getchar () ) x = x*10+c-'0';
}
int main (){
	int i, j, k, T;
	read (T);
	tot = 0;
	while ( T -- ){
		len = 0; memset ( first, 0, sizeof (first) );
		read (n); read (m); read (t);
		for ( i = 1; i <= n; i ++ ) rt[i] = 0, faa[i] = i, gs[i] = 1, dep[i] = 1;
		for ( i = 1; i <= n; i ++ ){
			read (na[i]);
			b[i] = na[i];
		}
		sort ( b+1, b+n+1 );
		bl = unique ( b+1, b+n+1 ) - (b+1);
		for ( i = 1; i <= n; i ++ ){
			na[i] = lower_bound ( b+1, b+bl+1, na[i] ) - b;
			change ( rt[i], 1, bl, na[i] );
		}
		for ( i = 1; i <= m; i ++ ){
			int x, y;
			read (x); read (y);
			int fx = ff (x), fy = ff (y);
			if ( gs[fx] > gs[fy] ) swap ( x, y );
			faa[fx] = fy; gs[fy] += gs[fx];
			fa[x][0] = y; dep[x] = dep[y]+1;
			dfs1 ( x, y );
			ins ( x, y ); ins ( y, x );
		}
		char c;
		int lastans = 0;
		while ( t -- ){
			c = getchar ();
			for ( ; c > 'Z' || c < 'A'; c = getchar () );
			if ( c == 'Q' ){
				int x, y;
				read (x); read (y); read (k);
				x ^= lastans; y ^= lastans; k ^= lastans;
				int lca = getlca ( x, y );
				lastans = getrank ( rt[x], rt[y], rt[fa[lca][0]], lca, 1, bl, k );
				printf ( "%d
", lastans );
			}
			else {
				int x, y;
				read (x); read (y);
				x ^= lastans; y ^= lastans;
				int fx = ff (x), fy = ff (y);
				if ( gs[fx] > gs[fy] ) swap ( x, y );
				faa[fx] = fy; gs[fy] += gs[fx];
				fa[x][0] = y; dep[x] = dep[y]+1;
				dfs1 ( x, y );
				ins ( x, y ); ins ( y, x );
			}
		}
		break;
	}
	return 0;
}