[BZOJ 3223]Tyvj 1729 文艺平衡树

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

题解

维护序列支持序列翻转可以用$splay$和$fhq-treap$实现。

其主要思想就是就是将这棵子树打上旋转标记，并交换左右儿子。

1 Splay

//It is made by Awson on 2017.12.18
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
using namespace std;
const int N = 100000;

int n, m, l, r;
struct Splay_tree {
    int pre[N+5], ch[N+5][2], key[N+5], rev[N+5], size[N+5], tot, root;
    void newnode(int &o, int keyy, int fa) {
    o = ++tot;
    key[o] = keyy, pre[o] = fa; size[o] = 1;
    ch[o][0] = ch[o][1] = rev[o] = 0;
    }
    void pushup(int o) {
    size[o] = size[ch[o][0]]+size[ch[o][1]]+1;
    }
    void pushdown(int o) {
    if (!o || !rev[o]) return;
    int ls = ch[o][0], rs = ch[o][1];
    rev[ls] ^= 1, rev[rs] ^= 1;
    swap(ch[ls][0], ch[ls][1]);
    swap(ch[rs][0], ch[rs][1]);
    rev[o] = 0;
    }
    void device(int &o, int fa, int l, int r) {
    if (l > r) return;
    int mid = (l+r)>>1;
    newnode(o, mid, fa);
    if (l == r) return;
    device(ch[o][0], o, l, mid-1);
    device(ch[o][1], o, mid+1, r);
    pushup(o);
    }
    void rotate(int o, int kind) {
    int p = pre[o];
    ch[p][!kind] = ch[o][kind], pre[ch[o][kind]] = p;
    ch[pre[p]][ch[pre[p]][1] == p] = o, pre[o] = pre[p];
    ch[o][kind] = p; pre[p]= o;
    pushup(p), pushup(o);
    }
    void splay(int o, int goal) {
    while (pre[o] != goal) {
        pushdown(pre[o]), pushdown(o);
        if (pre[pre[o]] == goal) rotate(o, ch[pre[o]][0] == o);
        else {
        int p = pre[o], kind = ch[pre[p]][0] == p;
        if (ch[p][kind] == o) rotate(o, !kind), rotate(o, kind);
        else rotate(p, kind), rotate(o, kind);
        }
    }
    if (!goal) root = o;
    }
    int get_node(int o, int rank) {
    pushdown(o);
    if (size[ch[o][0]]+1 == rank) return o;
    if (size[ch[o][0]] >= rank) return get_node(ch[o][0], rank);
    return get_node(ch[o][1], rank-(size[ch[o][0]]+1));
    }
    void reser(int l, int r) {
    int r1 = get_node(root, l), r2 = get_node(root, r);
    splay(r1, 0), splay(r2, r1);
    int p = ch[r2][0];
    swap(ch[p][0], ch[p][1]); rev[p] ^= 1;
    }
    void print(int o) {
    pushdown(o);
    if (ch[o][0]) print(ch[o][0]);
    if (key[o] != 1 && key[o] != n+2) printf("%d ", key[o]-1);
    if (ch[o][1]) print(ch[o][1]);
    }
}S;

void work() {
    scanf("%d%d", &n, &m);
    S.device(S.root, 0, 1, n+2);
    while (m--) {
    scanf("%d%d", &l, &r);
    S.reser(l, r+2);
    }
    S.print(S.root); printf("\n");
}
int main() {
    work();
    return 0;
}

2 fhq_Treap

对于无旋$Treap$，代码里废除了平衡参数$lev$的思想。在$merge$的时候直接随机出一个参数，来判断左接还是右接。

//It is made by Awson on 2017.12.19
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
using namespace std;
const int N = 100000;

int n, m, l, r;
struct fhq_Treap {
    int ch[N+5][2], key[N+5], size[N+5], rev[N+5], tot, root;
    void newnode(int &o, int keyy) {
    o = ++tot;
    ch[o][0] = ch[o][1] = rev[o] = 0;
    key[o] = keyy, size[o] = 1;
    }
    void pushup(int o) {
    size[o] = size[ch[o][0]]+size[ch[o][1]]+1;
    }
    void pushdown(int o) {
    if (!o || !rev[o]) return;
    int ls = ch[o][0], rs = ch[o][1];
    rev[ls] ^= 1, rev[rs] ^= 1;
    swap(ch[ls][0], ch[ls][1]);
    swap(ch[rs][0], ch[rs][1]);
    rev[o] = 0;
    }
    void device(int &o, int l, int r) {
    if (l > r) return;
    int mid = (l+r)>>1;
    newnode(o, mid);
    if (l == r) return;
    device(ch[o][0], l, mid-1);
    device(ch[o][1], mid+1, r);
    pushup(o);
    }
    void split(int o, int k, int &x, int &y) {
    if (!o) x = y = 0;
    else {
        pushdown(o);
        if (size[ch[o][0]]+1 <= k) x = o, split(ch[o][1], k-(size[ch[o][0]]+1), ch[o][1], y);
        else y = o, split(ch[o][0], k, x, ch[o][0]);
        pushup(o);
    }
    }
    int merge(int x, int y) {
    if (!x || !y) return x+y;
    pushdown(x), pushdown(y);    
    if (rand()%2) {
        ch[x][1] = merge(ch[x][1], y);
        pushup(x); return x;
    }else {
        ch[y][0] = merge(x, ch[y][0]);
        pushup(y); return y;
    }
    }
    void reser(int l, int r) {
    int r1, r2, r3;
    split(root, r, r2, r3);
    split(r2, l-1, r1, r2);
    rev[r2] ^= 1; swap(ch[r2][0], ch[r2][1]);
    root = merge(merge(r1, r2), r3);
    }
    void print(int o) {
    pushdown(o);
    if (ch[o][0]) print(ch[o][0]);
    printf("%d ", key[o]);
    if (ch[o][1]) print(ch[o][1]);
    }
}T;

void work() {
    srand(time(0));
    scanf("%d%d", &n, &m);
    T.device(T.root, 1, n);
    while (m--) {
    scanf("%d%d", &l, &r);
    T.reser(l, r);
    }
    T.print(T.root); printf("\n");
}
int main() {
    work();
    return 0;
}