NEERC 17 Problem I. Interactive Sort

Solution

當寫了兩倍正解的代碼使用了兩倍于正解的操作步數……

剛開始的想法是求出一個$b$，再求出一個$a$，依次輪換，分別維護當前知道確定值的$a,b$的有序序列，然后求$a$就在$b$的有序序列里二分，確定大致范圍，再暴力求出當前數的大小并更新$b$中每個數的上下界，求$b$同理。

這樣操作次數應該是$O(nlgn)$的，然而常數太大了……，需要接近$40W$次操作。

而正解則是上一個做法的一半。。。
我們依次用上述方法求出每一個$b$，可以發現顯然到最后$a$中的數的上下界相等，可以唯一確定一個$a$，所以大概只需要一半的操作步數了。

時間復雜度$O(n^2)$，代碼是$O(n^{2}lgn)$的，多一個$map$也沒慢多少。

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; } 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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