[BZOJ3223]Tyvj 1729 文艺平衡树

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

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

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;int read() {
int x = 0, f = 1; char c = getchar();
while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
return x * f;
}#define maxn 100010
struct Node {
int v, siz; bool rev;
Node() {}
Node(int _): v(_), rev(0) {}
} ns[maxn];
int rt, ToT, fa[maxn], ch[2][maxn];
void maintain(int o) {
ns[o].siz = 1;
for(int i = 0; i < 2; i++) if(ch[i][o])
ns[o].siz += ns[ch[i][o]].siz;
return ;
}
void build(int& o, int l, int r) {
if(l > r) return ;
int mid = l + r >> 1;
ns[o = ++ToT] = Node(mid);
if(l == r) return maintain(o);
build(ch[0][o], l, mid - 1); build(ch[1][o], mid + 1, r);
if(ch[0][o]) fa[ch[0][o]] = o;
if(ch[1][o]) fa[ch[1][o]] = o;
return maintain(o);
}
void pushdown(int o) {
for(int i = 0; i < 2; i++) if(ch[i][o])
ns[ch[i][o]].rev ^= ns[o].rev;
if(ns[o].rev) swap(ch[0][o], ch[1][o]), ns[o].rev = 0;
return ;
}
void rotate(int u) {
int y = fa[u], z = fa[y], l = 0, r = 1;
if(z) ch[ch[1][z]==y][z] = u;
if(ch[1][y] == u) swap(l, r);
fa[u] = z; fa[y] = u; fa[ch[r][u]] = y;
ch[l][y] = ch[r][u]; ch[r][u] = y;
maintain(y); maintain(u);
return ;
}
int S[maxn], top;
void splay(int u) {
int t = u; S[++top] = t;
while(fa[t]) t = fa[t], S[++top] = t;
while(top) pushdown(S[top--]);
while(fa[u]) {
int y = fa[u], z = fa[y];
if(z) {
if(ch[0][y] == u ^ ch[0][z] == y) rotate(u);
else rotate(y);
}
rotate(u);
}
return ;
}
int split(int u) {
if(!u) return 0;
splay(u);
int tmp = ch[1][u];
fa[tmp] = 0; ch[1][u] = 0;
maintain(u);
return tmp;
}
int merge(int a, int b) {
if(!a) return maintain(b), b;
if(!b) return maintain(a), a;
pushdown(a); while(ch[1][a]) a = ch[1][a], pushdown(a);
splay(a);
ch[1][a] = b; fa[b] = a;
return maintain(a), a;
}
int qkth(int o, int k) {
if(!o) return 0;
pushdown(o);
int ls = ch[0][o] ? ns[ch[0][o]].siz : 0;
if(k == ls + 1) return o;
if(k > ls + 1) return qkth(ch[1][o], k - ls - 1);
return qkth(ch[0][o], k);
}
int Find(int k) {
rt = 1; while(fa[rt]) rt = fa[rt];
return qkth(rt, k);
}
void Rev(int ql, int qr) {
int lrt = Find(ql - 1), mrt = Find(qr), rrt;
split(lrt); rrt = split(mrt);
ns[mrt].rev ^= 1;
mrt = merge(lrt, mrt); merge(mrt, rrt);
return ;
}int main() {
int n = read(), q = read();build(rt, 1, n);
while(q--) {
int l = read(), r = read();
Rev(l, r);
}for(int i = 1; i <= n; i++) {
int k = Find(i);
printf("%d ", ns[k].v);
}return 0;
}