Splay ---- 2018牛客多校第三场区间翻转搞区间位移或者 rope可持久化块状链表

题目链接

题目大意：

就是每次把牌堆中若干个连续的牌放到堆顶，问你最后牌的序列。

解题思路：

Splay 区间翻转的模板题：
对于一个区间[1,2,3,4,5,6,7,8][1,2,3,4,5,6,7,8][1,2,3,4,5,6,7,8]
你要把[3,4,5][3,4,5][3,4,5]拿到前面去

我们可以这么搞先把[1−5][1-5][1−5]翻转·一下
[5,4,3,2,1,6,7,8][5,4,3,2,1,6,7,8][5,4,3,2,1,6,7,8]
然后我们发现只要把[5,4,3][5,4,3][5,4,3]和[2,1][2,1][2,1]再翻转一遍就好了
[3,4,5,1,2,6,7,8][3,4,5,1,2,6,7,8][3,4,5,1,2,6,7,8]

AC code

#include <bits/stdc++.h>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x7fffffff
#define LLF 0x3f3f3f3f3f3f3f3f
#define f first
#define s second
#define endl '\n'
using namespace std;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {read(first);read(args...);
}
int n, m, arr[maxn];
struct node {#define ls(x) T[x].ch[0]#define rs(x) T[x].ch[1]#define fa(x) T[x].fa#define root T[0].ch[1]/*** 树形结构*/int fa;//父亲int ch[2];// 存储左右儿子的位置int sum;// 子树大小int val;  /*** * lazy 标记两个* */int rac;/*** 权值 最大值*/
}T[N];
void downdate(int x) {if(T[x].rac) {int &lc = T[x].ch[0], &rc = T[x].ch[1];swap(lc,rc);T[lc].rac ^= 1, T[rc].rac ^= 1;T[x].rac = 0;}
}void update(int x) {T[x].sum = T[ls(x)].sum + T[rs(x)].sum + 1;
}int ident(int x) {return T[fa(x)].ch[0] == x ? 0 : 1;}
void connect(int x, int fa, int how) {// 把x变成fa的how儿子T[fa].ch[how] = x;T[x].fa = fa;
}void rotate(int x) { // 把x转到fa位置int Y = fa(x), R = fa(Y);int YSON = ident(x), RSON = ident(Y);downdate(Y);downdate(x);connect(T[x].ch[YSON^1],Y,YSON), connect(Y,x,YSON^1), connect(x,R,RSON);update(Y);update(x);
}void splay(int x, int to) { // 把x转到to的位置to = fa(to);while(fa(x) != to) {int y = fa(x);if(T[y].fa == to) rotate(x);else if(ident(x) == ident(y)) rotate(y), rotate(x);else rotate(x), rotate(x);}
}
//.....................................................................
int find(int x) {int now = root;while (now) {downdate(now);if (x <= T[T[now].ch[0]].sum) now = T[now].ch[0];else {x -= T[T[now].ch[0]].sum + 1;if (!x) return now;now = T[now].ch[1];}}return 0;
}inline int BuildTree(int l,int r) {if(l > r) return 0;connect(BuildTree(l,mid-1),mid,0);connect(BuildTree(mid+1,r),mid,1);T[mid].val = mid;update(mid);return mid;
}void print(int now) {if(!now) return;downdate(now);if(T[now].ch[0]) print(T[now].ch[0]);if(now != 1 && now != n + 2) cout << now - 1 << " ";// cout << now << " ";if(T[now].ch[1]) print(T[now].ch[1]);
}inline void work(int l, int r) {int fl = find(l);int fr = find(r+2);splay(fl,root);splay(fr,T[root].ch[1]);T[T[fr].ch[0]].rac ^= 1;
}int main() {read(n,m);root = BuildTree(1,n+2);while(m --) {int l, r;cin >> l >> r;work(1,l+r-1);work(1,r);work(r+1,r+l-1);}print(root);return 0;
}

rope

Rope其主要是结合了链表和数组各自的优点，链表中的节点指向每个数据

块，即数组，并且记录数据的个数，然后分块查找和插入。在g++头文件中，< ext / rope >中有成型的块状链表，在using namespace

__gnu_cxx;空间中，其操作十分方便。

基本操作：
rope test;

test.push_back(x);//在末尾添加x

test.insert(pos,x);//在pos插入x

test.erase(pos,x);//从pos开始删除x个

test.copy(pos,len,x);//从pos开始到pos+len为止用x代替

test.replace(pos,x);//从pos开始换成x

test.substr(pos,x);//提取pos开始x个 x的默认值是1

test.at(x)/[x];//访问第x个元素

#include <bits/stdc++.h>
#include <ext/rope>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define LLF 0x3f3f3f3f3f3f3f3f
#define f first
#define s second
#define endl '\n'
using namespace std;
using namespace __gnu_cxx;
const int N = 2e6 + 10, mod = 1e9 + 9;
const int maxn = 500010;
const long double eps = 1e-5;
const int EPS = 500 * 500;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) {read(first);read(args...);
}int main() {IOS;int n, m;rope<int> q;cin >> n >> m;for(int i = 1; i <= n; ++ i) {q.push_back(i);}for(int i = 1; i <= m; ++ i) {int u, v;cin >> u >> v;q = q.substr(u-1,v) + q.substr(0,u-1) + q.substr(u+v-1,n-(u+v-1));}for(int i = 0; i < n; ++ i)cout << q[i] << " ";
}