标签：lex sse empty port ide *** pac long optimize
题意：一个数组，两种操作，1把1到r变成递增，2把1到r变成递减
解法：首先可以明确的是有效操作r肯定是递减的（因为不递减后面的操作会覆盖前面的操作），1,2然后肯定是交替的，因为如果不是交替的，那么结果不会变，所以是无效操作，
然后我们想办法解决这个有效操作，可以用splay打翻转标记
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> #define cd complex<double> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double eps=1e-6; const int N=200000+10,maxn=50+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int a[N]; struct Splay{ 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; 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(); } Node *root,*left,*right; void init(int sz) { null=new Node; null->s=0; root=new Node; root->v=a[1];root->flip=0; root->ch[0]=root->ch[1]=null; root->maintain(); Node* p; for(int i=2;i<=sz;i++) { p=new Node; p->v=a[i];p->s=p->flip=0; p->ch[0]=root,p->ch[1]=null; root=p; root->maintain(); } } void print(Node *o) { o->pushdown(); if(o->ch[0]!=null)print(o->ch[0]); printf("%d ",o->v); if(o->ch[1]!=null)print(o->ch[1]); } }sp; int main() { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++)scanf("%d",&a[i]); stack<pair<int,int> >s; for(int i=0;i<m;i++) { int t,r;scanf("%d%d",&t,&r); while(!s.empty()&&s.top().se<=r)s.pop(); if(s.empty())s.push(mp(t,r)); else if(s.top().fi!=t)s.push(mp(t,r)); } // while(!s.empty())printf("%d %d\n",s.top().fi,s.top().se),s.pop(); vector<pair<int,int> >v; while(!s.empty())v.pb(s.top()),s.pop(); reverse(v.begin(),v.end()); // for(int i=0;i<v.size();i++)printf("%d+++%d\n",v[i].fi,v[i].se); sort(a+1,a+v[0].se+1); // for(int i=1;i<=n;i++)printf("%d\n",a[i]); sp.init(n); for(int i=0;i<v.size();i++) { if(i==0&&v[i].fi==1)continue; int t=v[i].fi,r=v[i].se; if(r==n)sp.root->flip^=1; else { sp.split(sp.root,v[i].se,sp.left,sp.right); sp.left->flip^=1; sp.root=sp.Merge(sp.left,sp.right); } } sp.print(sp.root);puts(""); return 0; } /******************** 5 4 2 1 3 5 4 1 5 2 5 1 5 2 5 ********************/
也可以直接维护最后的区间，用栈保存答案
//#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-stack-protector") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") //#pragma GCC optimize("unroll-loops") #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair #define pb push_back #define pi acos(-1.0) #define ll long long #define vi vector<int> #define mod 1000000007 #define C 0.5772156649 #define ls l,m,rt<<1 #define rs m+1,r,rt<<1|1 #define pil pair<int,ll> #define pli pair<ll,int> #define pii pair<int,int> #define cd complex<double> #define ull unsigned long long #define base 1000000000000000000 #define fio ios::sync_with_stdio(false);cin.tie(0) using namespace std; const double g=10.0,eps=1e-12; const int N=200000+10,maxn=1000000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f; int a[N]; int main() { int n,m; scanf("%d%d",&n,&m); for(int i=1;i<=n;i++)scanf("%d",&a[i]); stack<pair<int,int> >s; for(int i=0;i<m;i++) { int t,r;scanf("%d%d",&t,&r); while(!s.empty()&&s.top().se<=r)s.pop(); if(s.empty())s.push(mp(t,r)); else if(s.top().fi!=t)s.push(mp(t,r)); } vector<pair<int,int> >v; while(!s.empty())v.pb(s.top()),s.pop(); reverse(v.begin(),v.end()); sort(a+1,a+v[0].se+1); // for(int i=1;i<=n;i++)printf("%d ",a[i]);puts(""); // for(int i=0;i<v.size();i++)printf("%d+++%d\n",v[i].fi,v[i].se); stack<int>st; int l=1,r=n,now=1; for(int i=0;i<v.size();i++) { int num=r-l+1-v[i].se; if(now==1) { for(int j=r;num!=0;j--,num--) st.push(a[j]); num=r-l+1-v[i].se; r=r-num; } else { for(int j=l;num!=0;j++,num--) st.push(a[j]); num=r-l+1-v[i].se; l+=num; } now=v[i].fi; } if(v[v.size()-1].fi==2) { for(int i=l;i<=r;i++) st.push(a[i]); } else { for(int i=r;i>=l;i--) st.push(a[i]); } while(!st.empty())printf("%d ",st.top()),st.pop(); return 0; } /*********************** 10 3 6 4 0 2 -3 7 8 -9 1 5 1 8 1 4 2 2 ***********************/
Codeforces Round #344 (Div. 2)C. Report
标签：lex sse empty port ide *** pac long optimize
原文地址：https://www.cnblogs.com/acjiumeng/p/8954596.html
