bzoj 3223: Tyvj 1729 文艺平衡树

3223: Tyvj 1729 文艺平衡树

Time Limit: 10 Sec Memory Limit: 128 MB

Submit: 2018 Solved: 1123

[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

平衡树

思路：

伸展树实现翻转

#include <iostream>
#include <cstdio>
using namespace std;

#define LS(x) node[(x)].ch[0]
#define RS(x) node[(x)].ch[1]

const int N = 1e5 + 5;

struct Splay {
struct Node {
int fa, ch[2];
bool rev;
int val, size;
void init(int _val) {
val = _val;
size = 1;
rev = ch[0] = ch[1] = 0;
}
} node
;

int root;

void up(int x) {
node[x].size = node[LS(x)].size + node[RS(x)].size + 1;
}

void rotate(int x, bool kind) {
int fx = node[x].fa;
int ffx = node[fx].fa;
node[fx].ch[!kind] = node[x].ch[kind];
node[node[x].ch[kind]].fa = fx;

node[x].ch[kind] = fx;
node[fx].fa = x;

node[ffx].ch[RS(ffx) == fx] = x;
node[x].fa = ffx;
up(fx);
}

void down(int x) {
if(x == 0) return ;
if(node[x].rev) {
if(LS(x)) node[LS(x)].rev ^= 1;
if(RS(x)) node[RS(x)].rev ^= 1;
swap(LS(x), RS(x));
node[x].rev = 0;
}
}

void splay(int x, int goal) {
while(node[x].fa != goal) {
int fx = node[x].fa;
int ffx = node[fx].fa;
down(ffx); down(fx); down(x);
bool rot_x = (LS(fx) == x);
bool rot_fx = (LS(ffx) == fx);
if(ffx == goal) rotate(x, rot_x);
else {
if(rot_x == rot_fx) rotate(fx, rot_fx);
else rotate(x, rot_x);
rotate(x, rot_fx);
}
}
up(x);
if(goal == 0) root = x;
}

int select(int pos) {
int u = root;
down(u);
while(node[LS(u)].size != pos) {
if(pos < node[LS(u)].size)
u = LS(u);
else {
pos -= node[LS(u)].size + 1;
u = RS(u);
}
down(u);
}
return u;
}

void rev(int L, int R) {
int u = select(L - 1), v = select(R + 1);
splay(u, 0); splay(v, u);
node[LS(v)].rev ^= 1;
}

int build(int L, int R) {
if(L > R) return 0;
if(L == R) return L;
int mid = (L + R) >> 1;
int r_L, r_R;
LS(mid) = r_L = build(L, mid - 1);
RS(mid) = r_R = build(mid + 1, R);
node[r_L].fa = node[r_R].fa = mid;
up(mid);
return mid;
}

void init(int n) {
node[0].init(0); node[0].size = 0;
node[1].init(0);
node[n + 2].init(0);
for(int i = 2; i <= n + 1; ++i)
node[i].init(i - 1);
root = build(1, n + 2);
}
void print(int);
} splay;

int ans
, tot;

void Splay :: print(int u) {
if(u == 0) return ;
down(u);
print(LS(u));
if(node[u].val)
ans[tot++] = node[u].val;
print(RS(u));
}

int main() {
int n, m;
scanf("%d%d", &n, &m);
splay.init(n);
for(int i = 0; i < m; ++i) {
int u, v;
scanf("%d%d", &u, &v);
splay.rev(u, v);
}
splay.print(splay.root);
for(int i = 0; i < tot; ++i) {
printf("%d ", ans[i]);
}
puts("");
return 0;
}