BZOJ1269 文本编辑器Editor

本文详细解析了BZOJ1269文本编辑器Editor问题,该问题是BZOJ1507的加强版,不仅增加了数据规模,还引入了反转操作。文章提供了一段C++代码实现,通过使用Splay树来解决这个问题,并展示了如何在Splay树中实现pushdown等关键操作。

BZOJ 1269 文本编辑器Editor


解析:BZOJ1507加强版,首先是数据加强,然后是多了一个reverse,似乎没什么,但是这意味着你需要加上一系列的pushdown······千万不要把pushdown位置搞错,千万不要多打,千万不要少打,不要作死。


Source:

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

inline void R(int &v)
{
	char c = 0;
	bool p = true;
	v = 0;
	while(!isdigit(c))
	{
		if(c == '-')
			p = false;
		c = getchar();
	}
	while(isdigit(c))
	{
		v = (v << 3) + (v << 1) + (c ^ '0');
		c = getchar();
	}
	if(!p)
		v = -v;
}

template
struct memorypool {
    T buf[size], *tail, *end, *st[size];
    int top;
    memorypool() : top(0), tail(buf), end(buf + size) {}
    inline T *alloc() {
        if (top) return st[--top];
        if (tail != end) return tail++;
        return new T;
    }
    inline void recycle(T *p) { 
        if (top > size) delete p;
        else st[top++] = p;
    }
};
const int MAXN = 1024 * 1024 * 4 ;
const int MAX = 1024 * 1024 * 4 ;

template
struct Splay
{
	enum Relation { L = 0, R = 1 };
	struct node
	{
		node *child[2], *parent, **root;
		int size;
		T value;
		bool reverse, bound;
		inline void init(node *parent, node **root, const T &value, bool bound = false)
		{
			this->parent = parent, this->root = root, this->value = value, this->bound = bound;
			this->reverse = false, this->size = 1, child[L] = child[R] = NULL;
		}
		inline void recycle(memorypool &pool)
		{
			if(child[L]) pool.recycle(child[L]);
			if(child[R]) pool.recycle(child[R]);
		}
		inline void pushdown()
		{
			if(reverse)
			{
				if(child[L]) child[L]->reverse ^= 1;
				if(child[R]) child[R]->reverse ^= 1;
				swap(child[L], child[R]), reverse = false;
			}
		}
		inline Relation relation() { return this == parent->child[L] ? L : R; }
		inline void maintain() { pushdown(), size = (child[L] ? child[L]->size : 0) + (child[R] ? child[R]->size : 0) + 1; }
		inline void rotate()
		{
			if(parent->parent) parent->parent->pushdown();
			parent->pushdown(), pushdown();
			Relation x = relation();
			node *oldparent = parent;
			if(oldparent->parent) oldparent->parent->child[oldparent->relation()] = this;
			parent = oldparent->parent, oldparent->child[x] = child[x ^ 1];
			if(child[x ^ 1]) child[x ^ 1]->parent = oldparent;
			child[x ^ 1] = oldparent, oldparent->parent = this, oldparent->maintain(), maintain();
			if(!parent) *root = this; 
		}
		inline void splay(node *targetparent = NULL)
		{
			while(parent != targetparent)
			{
				if(parent->parent == targetparent) rotate();
				else
				{
					parent->parent->pushdown(), parent->pushdown();
					if(parent->relation() == relation()) parent->rotate(), rotate();
					else rotate(), rotate();
				}
			}
		}
		inline int rank() { return child[L] ? child[L]->size : 0; }
		inline int pos() { return splay(), child[L] ? child[L]->size : 0; }
	} *root;
	memorypool  pool;
	Splay() : root(NULL) { buildbound(L), buildbound(R); }
	inline node *buildrange(const T *a, int l, int r, node *parent)
	{
		if(l > r) return NULL;
		register int mid = l + r >> 1;
		node *v = pool.alloc();
		v->init(parent, &root, a[mid - 1]);
		if(l != r) v->child[L] = buildrange(a, l, mid - 1, v), v->child[R] = buildrange(a, mid+1, r, v);
		return v->maintain(), v; 
	}
	inline void buildbound(Relation x)
	{
		node **v = &root, *parent = NULL;
		while(*v) parent = (*v), parent->size++, v = &parent->child[x];
		(*v) = pool.alloc(), (*v)->init(parent, &root, 0, true), (*v)->maintain();
	}
	inline node *select(int k)
	{
		k++;
		node *v = root;
		while(v->pushdown(), v->rank() + 1 != k) v = (v->rank() >= k ? v->child[L] : (k -= v->rank() + 1, v->child[R]));
		return v->splay(), v;
	}
	inline node *&select(int l, int r)
	{
		node *vl = select(l - 1), *vr = select(r + 1);
		return vl->splay(), vr->splay(vl), vr->child[L];
	}
	inline void erase(int l, int r)
	{
		node *&range = select(l, r);
		range->recycle(pool), pool.recycle(range), range = NULL;
		root->child[R]->maintain(), root->maintain();
	}
	inline void insert(int pos, const T *a, int n)
	{
		node *&range = select(pos + 1, pos);
		range = buildrange(a, 1, n, root->child[R]);
		root->child[R]->maintain(), root->maintain();
	}
	inline void reverse(int l, int r)
	{
		node *range = select(l ,r);
		range->reverse ^= 1;
	}
	inline int size() { return root->size - 2; }
};

Splay splay;

int main()
{
	static char type[10], a[MAX], c;
	static int x, pos, n;
	R(n);
	while(n--)
	{
		scanf("%s", type);
		if(type[0] == 'I') 
		{
			R(x);
			int cnt = 0;
			while(true)
			{
				c = getchar();
				if(c >= 32 && c <= 126)
					a[cnt++] = c;
				if(cnt == x) break;
			}
			splay.insert(pos, a, x);
		}
		else if(type[0] == 'M') R(x), pos = x;
		else if(type[0] == 'D') R(x), splay.erase(pos + 1, pos + x);
		else if(type[0] == 'R') R(x), splay.reverse(pos + 1, pos + x);
		else if(type[0] == 'G') cout << splay.select(pos + 1)->value << '\n';
		else if(type[0] == 'P') pos--;
		else if(type[0] == 'N') pos++;
	}
	return 0;
}  

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