【模板】文艺平衡树(Splay) 区间翻转 BZOJ 3223

您需要写一种数据结构(可参考题目标题),来维护一个有序数列,其中需要提供以下操作:翻转一个区间,例如原有序序列是5 4 3 2 1,翻转区间是[2,4]的话,结果是5 2 3 4 1 



N,M<=100000

Sample Output4 3 2 1 5 Hint


Input

第一行为n,m n表示初始序列有n个数,这个序列依次是(1,2……n-1,n)  m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n 

Output

 

输出一行n个数字,表示原始序列经过m次变换后的结果 

Sample Input5 3 1 3 1 3 1 4
 
Splay Tree ,和上道题一样,我们的kth 返回的序列的第k个数;
然后中序遍历即可得到我们翻转后的序列;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#pragma GCC optimize(2)
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)

inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

struct splay {
	int fa, ch[2], size;
	int lazy, rev, maxx, value;
}Sp[maxn];

int n, m, root, a[maxn];
void pushup(int rt) {
	Sp[rt].size = Sp[Sp[rt].ch[0]].size + Sp[Sp[rt].ch[1]].size + 1;
	Sp[rt].maxx = max(Sp[rt].value, max(Sp[Sp[rt].ch[0]].maxx, Sp[Sp[rt].ch[1]].maxx));
}

void pushdown(int rt) {
	if (Sp[rt].lazy) {
		if (Sp[rt].ch[0]) {
			Sp[Sp[rt].ch[0]].lazy += Sp[rt].lazy;
			Sp[Sp[rt].ch[0]].maxx += Sp[rt].lazy;
			Sp[Sp[rt].ch[0]].value += Sp[rt].lazy;
		}
		if (Sp[rt].ch[1]) {
			Sp[Sp[rt].ch[1]].lazy += Sp[rt].lazy;
			Sp[Sp[rt].ch[1]].maxx += Sp[rt].lazy;
			Sp[Sp[rt].ch[1]].value += Sp[rt].lazy;
		}
		Sp[rt].lazy = 0;
	}
	if (Sp[rt].rev) {
		if (Sp[rt].ch[0]) {
			Sp[Sp[rt].ch[0]].rev ^= 1;
			swap(Sp[Sp[rt].ch[0]].ch[0], Sp[Sp[rt].ch[0]].ch[1]);
		}
		if (Sp[rt].ch[1]) {
			Sp[Sp[rt].ch[1]].rev ^= 1;
			swap(Sp[Sp[rt].ch[1]].ch[0], Sp[Sp[rt].ch[1]].ch[1]);
		}
		Sp[rt].rev = 0;
	}
}

int id(int x) {
	return Sp[Sp[x].fa].ch[1] == x;
}
void link(int son, int fa, int k) {
	Sp[son].fa = fa; Sp[fa].ch[k] = son;
}

void rotate(int  x) {
	int y = Sp[x].fa;
	int z = Sp[y].fa;
	int yk = id(x);
	int zk = id(y);
	int b = Sp[x].ch[yk ^ 1];
	link(b, y, yk); link(y, x, yk ^ 1);
	link(x, z, zk);
	pushup(y); pushup(x);
}

void SPLAY(int x, int aim) {
	while (Sp[x].fa != aim) {
		int y = Sp[x].fa;
		int z = Sp[y].fa;
		if (z != aim)id(x) == id(y) ? rotate(y) : rotate(x);
		rotate(x);
	}
	if (aim == 0)root = x;
}

int kth(int k) {
	int now = root;
	while (1) {
		pushdown(now);
		int left = Sp[now].ch[0];
		if (Sp[left].size + 1 < k) {
			k -= Sp[left].size + 1; now = Sp[now].ch[1];
		}
		else if (Sp[left].size >= k)now = left;
		else return now;
	}
}

int build(int l, int r, int fa) {
	if (l > r)return 0;
	if (l == r) {
		Sp[l].fa = fa; Sp[l].maxx = Sp[l].value = a[l];
		Sp[l].size = 1; return l;
	}
	int mid = (l + r) >> 1;
	Sp[mid].ch[0] = build(l, mid - 1, mid);
	Sp[mid].ch[1] = build(mid + 1, r, mid);
	Sp[mid].value = a[mid];
	Sp[mid].fa = fa;
	pushup(mid);
	return mid;
}

int split(int l, int r) {
	l = kth(l); r = kth(r + 2);
	SPLAY(l, 0); SPLAY(r, l);
	return Sp[Sp[root].ch[1]].ch[0];
}

void upd(int l, int  r, int v) {
	int now = split(l, r);
	Sp[now].lazy += v; Sp[now].maxx += v; Sp[now].value += v;
	pushup(Sp[root].ch[1]); pushup(root);
}

void Reverse(int l, int r) {
	int now = split(l, r);
	Sp[now].rev ^= 1;
	swap(Sp[now].ch[0], Sp[now].ch[1]);
	pushup(Sp[root].ch[1]); pushup(root);
}
int query(int l, int  r) {
	return Sp[split(l, r)].maxx;
}

void dfs(int rt) {
	pushdown(rt);
	if (Sp[rt].ch[0])dfs(Sp[rt].ch[0]);
	if (Sp[rt].value != inf && Sp[rt].value != -inf) {
		cout << Sp[rt].value << ' ';
	}
	if (Sp[rt].ch[1])dfs(Sp[rt].ch[1]);
}

int main()
{
	//ios::sync_with_stdio(0);
	rdint(n); rdint(m);
	for (int i = 2; i <= n + 1; i++)a[i] = i - 1;
	a[1] = -inf; a[n + 2] = inf;
	root = build(1, n + 2, 0);
	while (m--) {
		int l, r; rdint(l); rdint(r);
		Reverse(l, r);
	}
	dfs(root); cout << endl;
    return 0;
}

 

EPFL - Fighting
posted @ 2018-12-05 19:59  NKDEWSM  阅读(184)  评论(0编辑  收藏  举报