BZOJ5099 POI2018Pionek

  假设确定了最终所得向量的方向,则应该选择所有在该方向上投影为正的向量。按极角序排序后这显然是一段连续区间。最终向量方向很难枚举,但对于某个向量,在其上投影为正的向量与其夹角范围是(-π/2,π/2),所以只要枚举所有极角差不超过π的极长区间就可以了。这里的区间不是向量区间而是极角区间,相当于一条过原点的直线在旋转,所以双指针移动时每个向量区间都要更新答案。为了方便倍长向量数组,注意这样对于增加的那一半,极角需要加上2π。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n;
const double PI=3.14159266;
ll ans;
struct vector
{
    int x,y;double angle;
    bool operator <(const vector&a) const
    {
        return angle<a.angle;
    }
    vector operator +(const vector&a) const
    {
        return (vector){x+a.x,y+a.y};
    }
    vector operator -(const vector&a) const
    {
        return (vector){x-a.x,y-a.y};
    }
    ll len(){return 1ll*x*x+1ll*y*y;}
}a[N<<1];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj5099.in","r",stdin);
    freopen("bzoj5099.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i].x=read(),a[i].y=read(),a[i].angle=atan2(a[i].x,a[i].y);
    sort(a+1,a+n+1);
    for (int i=n+1;i<=n*2;i++) a[i]=a[i-n],a[i].angle+=2*PI;
    vector cur=(vector){0,0};
    int x=0;
    for (int i=1;i<=n;i++)
    {
        while (x<n*2&&a[x+1].angle-a[i].angle<=PI) ans=max(ans,(cur=cur+a[++x]).len());
        ans=max(ans,(cur=cur-a[i]).len());
    }
    cout<<ans;
    return 0;
}

 

posted @ 2018-12-06 22:00  Gloid  阅读(216)  评论(0编辑  收藏  举报