BZOJ3223: Tyvj 1729 文艺平衡树

3223: Tyvj 1729 文艺平衡树

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 5918  Solved: 3535
[Submit][Status][Discuss]

Description

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 

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 Input

5 3

1 3

1 3

1 4

Sample Output

4 3 2 1 5

HINT

N,M<=100000

Source

【题解】

平衡树维护序列，大小顺序与节点内的权值无关，而是通过一开始钦定树的形态来钦定大小顺序

反转操作，把区间放到一颗子树内，显然子树根节点在区间中点，直接交换子树即可

/**************************************************************
    Problem: 3223
    User: 33511595
    Language: C++
    Result: Accepted
    Time:3040 ms
    Memory:13012 kb
****************************************************************/
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#include <string> 
#include <cmath> 
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
template <class T>
inline void swap(T& a, T& b)
{
    T tmp = a;a = b;b = tmp;
}
inline void read(int &x)
{
    x = 0;char ch = getchar(), c = ch;
    while(ch < '0' || ch > '9') c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    if(c == '-') x = -x;
}
 
const int INF = 0x3f3f3f3f;
const int MAXN = 500000 + 10;
 
int ch[MAXN][2], fa[MAXN], size[MAXN], root,tag[MAXN], data[MAXN];
int son(int x){return x == ch[fa[x]][1];}
void pushup(int rt)
{
    size[rt] = size[ch[rt][1]] + size[ch[rt][0]] + 1;
}
void pushdown(int rt)
{
    if(!tag[rt]) return; 
    int l = ch[rt][0], r = ch[rt][1];
    tag[l] ^= 1, tag[r] ^= 1;
    swap(ch[l][0], ch[l][1]);
    swap(ch[r][0], ch[r][1]);
    tag[rt] = 0;
}
void rotate(int x)
{
    int y = fa[x], z = fa[y], b = son(x), c = son(y), a = ch[x][!b];
    if(z) ch[z][c] = x; else root = x; fa[x] = z;
    if(a) fa[a] = y; ch[y][b] = a;
    ch[x][!b] = y, fa[y] = x;
    pushup(y), pushup(x);
}
void splay(int x, int i)
{
    while(fa[x] != i)
    {
        int y = fa[x], z = fa[y];
        if(z == i) rotate(x);
        else
            if(son(x) == son(y)) rotate(y), rotate(x);
            else rotate(x), rotate(x);
    }
}
int getkth(int rt, int x)
{
    pushdown(rt);int l = ch[rt][0];
    if(size[l] + 1 == x) return rt;
    else if(size[l] + 1 > x) return getkth(ch[rt][0], x);
    else return getkth(ch[rt][1], x - size[l] - 1);
}
void turn(int l, int r)
{
    l = getkth(root, l - 1), r = getkth(root, r + 1);
    splay(l, 0);splay(r, l);
    int now = ch[r][0];
    tag[now] ^= 1;
    swap(ch[now][0], ch[now][1]);
    pushdown(now);
}
int n, m, tmp1, tmp2;
void put(int x)
{
    if(!x) return;
    pushdown(x);
    put(ch[x][0]);
    if(x != 1 && x != n + 2)printf("%d ", data[x]);
    put(ch[x][1]);
    return;
}
int main()
{
    read(n), read(m);
    root = 1;
    for(int i = 1;i <= n + 1;++ i) ch[i][1] = i + 1, fa[i + 1] = i, size[i] = n + 3 - i;
    size[n + 2] = 1;
    for(int i = 2;i <= n + 1;++ i) data[i] = i - 1;
    for(int i = 1;i <= m;++ i)
    {
        read(tmp1), read(tmp2);
        turn(tmp1 + 1, tmp2 + 1);
    }
    put(root);
    return 0;
}