BZOJ 5099([POI2018]Pionek-极角排序)

题意：有n个向量，每个向量可以选择取或者不取，要求选一些向量，使得它们和最大。
假设最终答案的方向向量已知，显然只要取和它点积为正的向量就行。
维护一条过原点的直线，让它转一周，并维护这条直线上方的向量，将它们全取。

#include<bits/stdc++.h> 
using namespace std;
#define
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma
#define
#define
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
struct point{
    ll fi,se;
    double ang;
}p[MAXN];
int cmp(point a,point b) {
    return a.ang<b.ang;
}
const double PI=acos(-1);
int main()
{
//  freopen("bzoj5099.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read(),m=0;
    For(i,n){
        ll x=read(),y=read();
        if (x*x+y*y) p[++m]=point{x,y,atan2(y,x)};
    }
    n=m;
    sort(p+1,p+n+1,cmp);
    ll ans=0;
    ll sx=0,sy=0;
    For(i,n) {
        p[n+i]=p[i]; p[n+i].ang+=2*PI;
    }
    for(int i=1,j=1;i<=n;i++) {
        while(j<n+i && p[j].ang-p[i].ang<=PI+1e-13) {
            sx+=p[j].fi;
            sy+=p[j].se;
            gmax(ans, sx*sx+sy*sy); 
            ++j;
        }
        sx-=p[i].fi;    sy-=p[i].se;
        gmax(ans, sx*sx+sy*sy); 
    }
    cout<<ans<<endl;
    return 0;
}