CF1158D. Beautiful Array
Solution
构造
对于所有点(xi,yi)(x_i,y_i)(xi,yi)选择yiy_iyi最小的点作为起点，每次考虑下一步若是LLL，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(n2)O(n^2)O(n2)
#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 llusing 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;
}
