CF1158D. Beautiful Array

Solution

构造

对于所有点$(x_i,y_i)$选择$y_i$最小的点作为起点，每次考虑下一步若是$L$，则往最右边（与当前线段夹角最大）的点走，否则往最左边的点走。

时间复杂度$O(n^2)$

#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 second #define int ll using 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=200005; const int INF=0x3f3f3f3f; /*--------------------------------------------------------------------*/ inline ll read() { ll f=1,x=0; char c=getchar(); while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); } while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); } return x*f; } PR a[MAXN]; char st[MAXN]; int n,flag[MAXN],Ans[MAXN]; int cross(PR x,PR y,PR z) { return (y.fi-x.fi)*(z.se-x.se)-(y.se-x.se)*(z.fi-x.fi); } signed main() { n=read(); int mn=1; for (int i=1;i<=n;i++) a[i].fi=read(),a[i].se=read(),a[i].se<a[mn].se?mn=i:0; flag[mn]=1,Ans[1]=mn; scanf("%s",st+1); for (int i=2,lst=0;i<n;i++) { if (st[i-1]=='L') { int mx=0; for (int j=1;j<=n;j++) if (!flag[j]&&(!mx||cross(a[Ans[i-1]],a[mx],a[j])<0)) mx=j; Ans[i]=mx,flag[mx]=1; } else { int mx=0; for (int j=1;j<=n;j++) if (!flag[j]&&(!mx||cross(a[Ans[i-1]],a[mx],a[j])>0)) mx=j; Ans[i]=mx,flag[mx]=1; } } for (int i=1;i<n;i++) printf("%d ",Ans[i]); for (int i=1;i<=n;i++) if (!flag[i]) printf("%d\n",i); return 0; }