NEERC 17 Problem I. Interactive Sort

Solution

当写了两倍正解的代码使用了两倍于正解的操作步数……

刚开始的想法是求出一个 $b$ ，再求出一个 $a$ ，依次轮换，分别维护当前知道确定值的 $a, b$ 的有序序列，然后求 $a$ 就在 $b$ 的有序序列里二分，确定大致范围，再暴力求出当前数的大小并更新 $b$ 中每个数的上下界，求 $b$ 同理。

这样操作次数应该是 $O (n l g n)$ 的，然而常数太大了……，需要接近 $40 W$ 次操作。

而正解则是上一个做法的一半。。。
我们依次用上述方法求出每一个 $b$ ，可以发现显然到最后 $a$ 中的数的上下界相等，可以唯一确定一个 $a$ ，所以大概只需要一半的操作步数了。

时间复杂度 $O(n^2)$ ，代码是 $O(n^2lgn)$ 的，多一个 $m a p$ 也没慢多少。

Code one

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
int example_a[MAXN],example_b[MAXN];
inline int read() { int x; scanf("%d",&x); return x; }
inline char get_ch() { char c=' '; while (c!='<'&&c!='>') scanf("%c",&c); return c; }
inline char get_chxy(int x,int y) { return example_a[x]<example_b[y]?'<':'>'; }map<PR,int> Map;
int la[MAXN],lb[MAXN],ra[MAXN],rb[MAXN],a[MAXN],b[MAXN],Ansa[MAXN],Ansb[MAXN],ida[MAXN],idb[MAXN],n,Num=0;int get(int x,int y)
{if (Map.count(MP(x,y))) return Map[MP(x,y)];printf("? %d %d\n",x,y),fflush(stdout);return Map[MP(x,y)]=(get_ch()=='>');
}PR geta(int x)
{int l=1,r=x,L,R;while (l<r){int mid=(l+r+1)>>1;if (get(x,idb[mid])) l=mid;else r=mid-1;}if (!get(x,idb[l])) L=1,R=b[l];else L=b[l],R=(l+1>x?n:b[l+1]);return MP((L&1)?L+1:L,(R&1)?R-1:R);
}
void solvea(int t)
{int num=0;PR x=geta(t);
//	Num=0;for (int j=1;j<=(n+1)>>1;j++){if (rb[j]<x.fi) num++;if (lb[j]<x.se&&x.fi<rb[j]) num+=get(t,j),Num++;}
//	cout<<"AQueryNum:"<<Num<<endl;Ansa[t]=a[t]=num<<1,ida[t]=t;for (int j=1;j<=(n+1)>>1;j++)if (Map.count(MP(t,j))) {int p=Map[MP(t,j)];if (!p) upmax(lb[j],Ansa[t]+1);else upmin(rb[j],Ansa[t]-1);}while (t&&a[t]<a[t-1]) swap(a[t],a[t-1]),swap(ida[t],ida[t-1]),t--;
}PR getb(int x)
{int l=1,r=x-1,L,R;while (l<r){int mid=(l+r)>>1;if (get(ida[mid],x)) r=mid;else l=mid+1;}if (x==1) L=1,R=n;else if (!get(ida[r],x)) L=a[r],R=n;else L=(r==1?1:a[r-1]),R=a[r];return MP((!(L&1))?L+1:L,(!(R&1))?R-1:R);
}
void solveb(int t)
{int num=0;PR x=getb(t);
//	Num=0;for (int j=1;j<=n>>1;j++){if (ra[j]<x.fi) num++;if (la[j]<x.se&&x.fi<ra[j]) num+=(!get(j,t)),Num++;}
//	cout<<"BQueryNum:"<<Num<<" "<<x.fi<<" "<<x.se<<endl;Ansb[t]=b[t]=num<<1|1,idb[t]=t;for (int j=1;j<=n>>1;j++)if (Map.count(MP(j,t))) {int p=Map[MP(j,t)];if (!p) upmin(ra[j],Ansb[t]-1);else upmax(la[j],Ansb[t]+1);}while (t&&b[t]<b[t-1]) swap(b[t],b[t-1]),swap(idb[t],idb[t-1]),t--;
}
signed main()
{n=read();srand(time(0));for (int i=1;i<=n>>1;i++) example_a[i]=i*2;for (int i=1;i<=(n+1)>>1;i++) example_b[i]=i*2-1;random_shuffle(example_a+1,example_a+(n>>1)+1);random_shuffle(example_b+1,example_b+((n+1)>>1)+1);for (int i=1;i<=n/2;i++) la[i]=2,ra[i]=n-(n&1);for (int i=1;i<=(n+1)/2;i++) lb[i]=1,rb[i]=n-((n&1)^1);for (int i=1,l,r;i<=n/2;i++) solveb(i),solvea(i);if (n&1) solveb((n+1)>>1);putchar('!');for (int i=1;i<=n/2;i++) printf(" %d",Ansa[i]);for (int i=1;i<=(n+1)/2;i++) printf(" %d",Ansb[i]);fflush(stdout);return 0;
}

Code two

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
int example_a[MAXN],example_b[MAXN];
inline int read() { int x; scanf("%d",&x); return x; }
inline char get_ch() { char c=' '; while (c!='<'&&c!='>') scanf("%c",&c); return c; }
inline char get_chxy(int x,int y) { return example_a[x]<example_b[y]?'<':'>'; }map<PR,int> Map;
int la[MAXN],ra[MAXN],a[MAXN],b[MAXN],Ansb[MAXN],ida[MAXN],f[MAXN],n,Num=0;int get(int x,int y)
{if (Map.count(MP(x,y))) return Map[MP(x,y)];printf("? %d %d\n",x,y),fflush(stdout);return Map[MP(x,y)]=(get_ch()=='>');
}
PR getb(int x)
{int num=0;for (int i=1;i<=n>>1;i++) f[i]=0;for (int i=1;i<=n>>1;i++) f[la[i]>>1]=i;for (int i=1;i<=n>>1;i++) if (f[i]) ida[++num]=f[i],a[num]=la[f[i]],b[num]=ra[f[i]];int l=1,r=num,L,R;while (l<r){Num++;int mid=(l+r)>>1;if (get(ida[mid],x)) r=mid;else l=mid+1;}if (!get(ida[r],x)) L=a[r],R=n;else L=(r==1?1:a[r-1]),R=b[r];return MP((!(L&1))?L+1:L,(!(R&1))?R-1:R);
}
void solveb(int t)
{int num=0;PR x=getb(t);for (int j=1;j<=n>>1;j++){if (ra[j]<x.fi) num++;if (la[j]<x.se&&x.fi<ra[j]) num+=(!get(j,t)),Num++;}Ansb[t]=num<<1|1;for (int j=1;j<=n>>1;j++)if (Map.count(MP(j,t))) {int p=Map[MP(j,t)];if (!p) upmin(ra[j],Ansb[t]-1);else upmax(la[j],Ansb[t]+1);}
}
signed main()
{n=read();if (n==1) { printf("! 1\n"); return 0; }for (int i=1;i<=n/2;i++) la[i]=2,ra[i]=n-(n&1);for (int i=1,l,r;i<=n/2;i++) solveb(i);if (n&1) solveb((n+1)>>1);putchar('!');for (int i=1;i<=n/2;i++) printf(" %d",la[i]);for (int i=1;i<=(n+1)/2;i++) printf(" %d",Ansb[i]);fflush(stdout);return 0;
}