hdu6127 Hard challenge

http://acm.hdu.edu.cn/showproblem.php?pid=6127

每条线都能把点集分成两部分A和B，每两个点之间都有线段，所以穿过的线段权值和也就是A点集中点的权值和乘以B点集中点的权值和

极角排序，把y轴下面的点按原点中心对称过来，然后和y轴上面的点相反处理，标记一下就可以

#include<bits/stdc++.h>#define N 50006using namespace std;const double pi=acos(-1.0);struct point{long long x,y,v;bool flag;}a[N];bool cmp(point x,point y){double xx,yy;xx=atan2((double)x.y,(double)x.x);yy=atan2((double)y.y,(double)y.x);return xx<yy;}int main(){int t,n;long long sum1=0,sum2=0;scanf("%d",&t);while(t--){sum1=0,sum2=0;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lld%lld%lld",&a[i].x,&a[i].y,&a[i].v);if(a[i].y<0){sum1+=a[i].v;a[i].x*=-1;a[i].y*=-1;a[i].flag=false;}else a[i].flag=true,sum2+=a[i].v;}sort(a,a+n,cmp);long long ans=sum1*sum2;for(int i=0;i<n;i++){//cout<<a[i].x<<" "<<a[i].y<<endl;if(a[i].flag){sum2-=a[i].v;sum1+=a[i].v;}else{sum1-=a[i].v;sum2+=a[i].v;}ans=max(ans,sum1*sum2);}cout<<ans<<endl;}return 0;}