题意:一个值1到n的数组,一种(多次)操作把l到r的区间反转,然后放到数组尾部
题解:裸的splay,用区间合并和区间分割,反转用lazy标记+pushdown就好了
#include<bits/stdc++.h> #include<ext/rope> #define fi first #define se second #define mp make_pair #define pb push_back #define pii pair<int,int> #define C 0.5772156649 #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 using namespace std; using namespace __gnu_cxx; const double g=10.0,eps=1e-7; const int N=100000+10,maxn=1000000+10,inf=0x3f3f3f; struct Node{ Node* ch[2]; int v; int s; int flip; int cmp(int x)const{ int d = x - ch[0]->s; if(d==1)return -1; return d<=0 ? 0:1; } void maintain() { s = 1 + ch[0]->s + ch[1]->s; } void pushdown() { if(flip)//类似于线段树的lazy标记 { flip=0; swap(ch[0],ch[1]); ch[0]->flip = !(ch[0]->flip); ch[1]->flip = !(ch[1]->flip); } } }; Node* null = new Node(); void Rotate(Node* &o,int d) { Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain();k->maintain(); o = k; } void splay(Node* &o,int k) { o->pushdown(); int d = o->cmp(k); if(d==1)k -= o->ch[0]->s + 1;//利用二叉树性质 if(d!=-1) { Node* p = o->ch[d]; p->pushdown(); int d2 = p->cmp(k); int k2 = (d2==0 ? k:k-p->ch[0]->s-1); if(d2!=-1) { splay(p->ch[d2],k2); if(d==d2)Rotate(o,d^1); else Rotate(o->ch[d],d); } Rotate(o,d^1); } } Node* Merge(Node* left,Node* right) { splay(left,left->s);//把排名最大的数splay到根 left->ch[1] = right; left->maintain(); return left; } void split(Node* o,int k,Node* &left,Node* &right) { splay(o,k);//把排名为k的节点splay到根,右侧子树所有节点排名比k大,左侧小 right = o->ch[1]; o->ch[1] = null; left = o; left->maintain(); } struct SplayTree{ int n; Node seq[N]; Node* root; Node* build(int sz) { if(sz==0)return null; Node* l=build(sz/2); Node* o=&seq[++n]; o->v=n; o->ch[0]=l; o->ch[1]=build(sz-sz/2-1); o->s = o->flip = 0; o->maintain(); return o; } void init(int sz) { n=0; null->s = 0; root = build(sz); } }; vector<int>ans; void print(Node* o) { if(o!=null) { o->pushdown(); print(o->ch[0]); ans.pb(o->v); print(o->ch[1]); } } void debug(Node* o) { if(o!=null) { o->pushdown(); debug(o->ch[0]); cout<<o->v<<endl; debug(o->ch[1]); } } SplayTree ss; int main() { int n,m; scanf("%d%d",&n,&m); ss.init(n+1); // debug(ss.root); while(m--) { int a,b; scanf("%d%d",&a,&b); Node *o,*left,*mid,*right; split(ss.root,a,left,o);//把ab整体右移一位,保证不会出现0 split(o,b-a+1,mid,right); mid->flip^=1; //把left+mid+right变成left+right+mid(fliped) ss.root = Merge(Merge(left,right),mid); } print(ss.root); for(int i=1;i<ans.size();i++) printf("%d\n",ans[i]-1); return 0; } /************ ************/
原文:http://www.cnblogs.com/acjiumeng/p/7748488.html