Problem

先告诉你每个盒子在哪个盒子的内部
接下来有M个操作:
可以把一个盒子及里面的盒子移到另外一个盒子的内部
或者询问你某个盒子最外面的盒子是哪个

Solution

首先可以建成一个图,然后先dfs一遍,用dfs序加入多棵splay
然后移动操作就是区间切割,询问操作就是这棵splay中最小的值

Notice

注意,因为有多棵splay,用0这个虚点连接起来。所以细节非常难处理

Code

#pragma GCC optimize(2)
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 100000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int Left[N + 5], Right[N + 5], T[N * 2 + 5], head[N + 5];
int point, num, root, time;
struct Node
{
	int vet, next;
}edge[N + 5];
void add(int u, int v)
{
	edge[++num].vet = v;
	edge[num].next = head[u];
	head[u] = num;
}
struct node
{
	int val[N + 5], parent[N + 5], son[2][N + 5];
	void Newnode(int &u, int from, int v)
	{
		u = ++point;
		val[u] = v, parent[u] = from;
		son[0][u] = son[1][u] = 0;
		if (v > 0) Left[v] = u;
		else Right[-v] = u;
	}
	void Build(int &u, int l, int r, int from)
	{
		int mid = (l + r) >> 1;
		Newnode(u, from, T[mid]);
		if (l < mid) Build(son[0][u], l, mid - 1, u);
		if (r > mid) Build(son[1][u], mid + 1, r, u);
	}
	
	void Rotate(int x)
	{
		int y = parent[x], z = parent[y];
		int l = (son[1][y] == x), r = l ^ 1;
		if (z != 0) son[son[1][z] == y][z] = x;
		parent[x] = z;
		parent[son[r][x]] = y, son[l][y] = son[r][x];
		parent[y] = x, son[r][x] = y;
	}
	void Splay(int x, int rt)
	{
		if (x == rt)  return;
		while (parent[x] != rt)
		{
			int y = parent[x], z = parent[y];
			if (z != rt)
				if ((son[0][y] == x) ^ (son[0][z] == y)) Rotate(x);
				else Rotate(y);
			Rotate(x);
		}
	}
	
	int Find_pre(int x)
	{
		x = son[0][x];
		while (son[1][x]) x = son[1][x];
		return x;
	}
	int Find_suf(int x)
	{
		x = son[1][x];
		while (son[0][x]) x = son[0][x];
		return x;
	}
	
	int Query(int x)
	{
		x = Left[x];
		Splay(x, 0);
		while (son[0][x]) x = son[0][x];
		return val[x];
	}
	void Move(int x, int y)
	{
		if (x == y) return;
		int xx = Left[x], yy = Right[x];
		Splay(xx, 0);
		Splay(yy, xx);
		int xxx = son[0][xx], yyy = son[1][yy], ttt = Find_pre(xx);
		son[0][xx] = 0, son[1][yy] = 0;
		parent[xxx] = parent[yyy] = 0;
		if (ttt) 
		{
			son[1][ttt] = yyy;
			parent[yyy] = ttt;
		}
		if (y == 0) return;
		if (Query(y) == x)
		{
			son[0][xx] = xxx, son[1][yy] = yyy;
			parent[xxx] = xx, parent[yyy] = yy;
			son[1][ttt] = 0; return;
		}
		Splay(Left[y], 0);
		ttt = Find_suf(Left[y]);
		Splay(ttt, Left[y]);
		son[0][son[1][Left[y]]] = xx;
		parent[xx] = son[1][Left[y]];
	}
}splay_tree;
void dfs(int u)
{
	T[time++] = u;
	travel(i, u)
	{
		int v = edge[i].vet;
		dfs(v);
	}
	T[time++] = -u;
}

int sqz()
{
	int n, flag = 0;
	while (~scanf("%d", &n))
	{
		if (flag) puts("");
		else flag = 1;
		point = num = time = 0;
		memset(head, 0, sizeof head);
		rep(i, 1, n)
		{
			int x = read();
			add(x, i);
		}
		dfs(0);
		int now = 0, last = 1;
		rep(i, 1, 2 * n)
		{
			if (T[i] > 0) now++;
			else now--;
			if (!now)
			{
				splay_tree.Build(root, last, i, 0);
				last = i + 1;
			}
		}
		int q = read(); char st[10];
		while(q--)
		{
			scanf("%s", st);
			if (st[0] == 'Q')
			{
				int x = read();
				printf("%d\n", splay_tree.Query(x));
			}
			else
			{
				int x = read(), y = read();
				splay_tree.Move(x, y);
			}
		}
	}
	return 0;
}
posted on 2017-10-06 12:44  WizardCowboy  阅读(170)  评论(0编辑  收藏  举报