2012 ACM/ICPC 杭州赛区 A题 HDU 4453 Looploop 伸展树

本文介绍了一种使用伸展树解决特定问题的算法实现。文章详细解释了如何通过伸展树来处理区间翻转、区间加操作及插入删除等操作，并提供了完整的C++代码示例。

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这题比较裸，一看就是伸展树。就是有些小细节要注意，为了避免反转时，指针变化，我是直接先把前面一段接在后面，这样就不会出现指针变化的问题。敲好以后一直TLE，调了半天，发现添加的时候我既然还是用n来添加= =（n表示现在的结点数，最后添加一个变量tn表示到现在为止出现过的结点数）。

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

using namespace std;
const int N = 200007;
#define KT ch[ch[root][1]][0]

int root, ch[N][2], fa[N], sz[N];
int n, m, k1, k2, id, tn;
bool rev[N];
int sum[N], num[N];

void build()
{
	fa[0] = fa[n + 1] = 0;
	sz[n + 1] = n + 2;
	ch[n + 1][1] = 1;
	fa[1] = n + 1;
	sz[1] = n + 1;
	for(int i = 2; i <= n; i++)
	{
		fa[i] = i - 1;
		ch[i -  1][1] = i;
		sz[i]= n + 2 - i;
	}
	fa[n + 2] = n;
	sz[n + 2] = 1;
	ch[n][1] = n + 2;
}

void init()
{
	sz[0] = 0;
	memset(ch, 0, sizeof(ch));
	num[0] = num[n + 1] = num[n + 2] = 0;
	for(int i =1; i <= n; i++)
		scanf("%d", num + i);
	memset(rev, false, sizeof(rev));
	memset(sum, 0, sizeof(sum));
	root = n + 1;
	id = 2;
	build();
}

inline void pushup(int x)
{
	sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
}

inline void pushdown(int x)
{
	if(rev[x])
	{
		rev[ch[x][0]] = !rev[ch[x][0]];
		rev[ch[x][1]] = !rev[ch[x][1]];
		swap(ch[x][0], ch[x][1]);
		rev[x] = !rev[x];
	}
	if(sum[x])
	{
		sum[ch[x][0]] += sum[x];
		sum[ch[x][1]] += sum[x];
		num[ch[x][0]] += sum[x];
		num[ch[x][1]] += sum[x];
		sum[x] = 0;
	}
}

inline void rotate(int x, bool f)
{
	int y = fa[x];
	int z = fa[y];
	pushdown(y);
	pushdown(x);
	ch[y][!f] = ch[x][f];
	fa[ch[x][f]] = y;
	fa[x] = z;
	if(z)
		ch[z][ch[z][1] == y] = x;
	ch[x][f] = y;
	fa[y] = x;
	pushup(y);
}

void splay(int x, int g)
{
	int y = fa[x];
	pushdown(x);
	while(y != g)
	{
		int z = fa[y];
		bool f = (ch[y][0] == x);
		if(z != g && f == (ch[z][0] == y))
			rotate(y, f);
		rotate(x, f);
		y = fa[x];
	}
	pushup(x);
	if(g == 0)
		root = x;
}

void rotateto(int k, int g)
{
	int x = root;
	pushdown(x);
	while(sz[ch[x][0]] != k)
	{
		if(k < sz[ch[x][0]])
			x = ch[x][0];
		else
		{
			k -= sz[ch[x][0]] + 1;
			x = ch[x][1];
		}
		pushdown(x);
	}
	splay(x, g);
}

void lin(int b)
{
	rotateto(0, 0);
	b -= n + 1;
	rotateto(b + 1, root);
	int temp = KT;
	KT = 0;
	sz[ch[root][1]] -= sz[temp];
	sz[root] -= sz[temp];
	id = id - sz[temp];
	fa[temp] = 0;
	rotateto(n - sz[temp], 0);
	rotateto(n + 1 - sz[temp], root);
	KT = temp;
	fa[temp] = ch[root][1];
	sz[ch[root][1]] += sz[temp];
	sz[root] += sz[temp];
}

void add(int c)
{
	int b = id + k2 - 1;
	if(b > n + 1)
		lin(b);
	int a = id;
	b = id + k2 - 1;
	rotateto(a - 2, 0);
	rotateto(b, root);
	sum[KT] += c;
	num[KT] += c;
}

void reve()
{
	int b = id + k1 - 1;
	if(b > n + 1)
		lin(b);
	b = id + k1 - 1;
	int a = id;
	rotateto(a - 2, 0);
	rotateto(b, root);
	rev[KT] = !rev[KT];
}

void insert(int c)
{
	rotateto(id - 1, 0);
	rotateto(id, root);
	n++;
	tn++;
	num[tn + 2] = c;
	sum[tn + 2] = rev[tn + 2] = 0;
	sz[tn + 2] = 1;
	ch[tn + 2][1] = ch[tn + 2][0] = 0;
	fa[tn + 2] = ch[root][1];
	KT = tn + 2;
	sz[ch[root][1]]++;
	sz[root]++;
}

void del()
{
	rotateto(id - 2, 0);
	rotateto(id, root);
	fa[KT] = 0;
	KT = 0;
	sz[ch[root][1]]--;
	sz[root]--;
	if(id == n + 1)
		id = 2;
	n--;
}

void move(int a)
{
	if(a == 1)
	{
		if(id == 2)
			id = n + 1;
		else
			id--;
	}
	else
	{
		if(id == n + 1)
			id = 2;
		else
			id++;
	}
}

int que()
{
	rotateto(id - 2, 0);
	rotateto(id, root);
	return num[KT];
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int C = 1;
	while(scanf("%d%d%d%d", &n, &m, &k1, &k2), n || m || k1 || k2)
	{
		init();
		tn = n;
		printf("Case #%d:\n", C++);
		while(m--)
		{
			char s[20];
			scanf("%s", s);
			int a;
			if(s[0] == 'a')
			{
				scanf("%d", &a);
				add(a);
			}
			else if(s[0] == 'r')
				reve();
			else if(s[0] == 'i')
			{
				scanf("%d", &a);
				insert(a);
			}
			else if(s[0] == 'd')
				del();
			else if(s[0] == 'm')
			{
				scanf("%d", &a);
				move(a);
			}
			else
				printf("%d\n", que());
		}
	}
	return 0;
}