HDU 6127 Hard challenge（几何）

Hard challenge

Problem Description

There are n points on the plane, and the ith points has a value vali, and its coordinate is (xi,yi). It is guaranteed that no two points have the same coordinate, and no two points makes the line which passes them also passes the origin point. For every two points, there is a segment connecting them, and the segment has a value which equals the product of the values of the two points. Now HazelFan want to draw a line throgh the origin point but not through any given points, and he define the score is the sum of the values of all segments that the line crosses. Please tell him the maximum score.

Input

The first line contains a positive integer T(1≤T≤5), denoting the number of test cases.
For each test case:
The first line contains a positive integer n(1≤n≤5×104).
The next n lines, the ith line contains three integers xi,yi,vali(|xi|,|yi|≤109,1≤vali≤104).

Output

For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input

2
2
1 1 1
1 -1 1
3
1 1 1
1 -1 10
-1 0 100

Sample Output

1
1100

Source

2017 Multi-University Training Contest - Team 7

题意：有n个点，给出每个点的坐标和value，相互连接，形成n*(n-1)/2条线段，每条线段的value等于两端点的value相乘。找出一个过原点的直线，使得与其相交的线段value值和最大。

题解：对于一条直线，线段权值和实际上就等于其两边点权和的乘积。把所有点按斜率排序，从大到小遍历扫描。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define INF 1e9+10
#define MAXN 305
using namespace std;
int n;
struct node
{
    double x;
    double y;
    double k;
    ll v;
}a[50005],b[50005];
int ca;
int cb;
bool cmp(node f,node g)
{
    return f.k>g.k;
}
int main()
{

    int t;
    scanf("%d",&t);

    for(int w=1;w<=t;w++)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        ca=0;cb=0;
        scanf("%d",&n);
        double x,y;
        ll z;
        ll sum1=0;
        ll sum2=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lld",&x,&y,&z);
            if(x<0||(x==0&&y<0))
            {
                b[cb].x=x;
                b[cb].y=y;
                b[cb].v=z;
                if(x==0&&y<0) b[cb].k=INF;
                else if(x==0&&y>0) b[cb].k=INF;
                else b[cb].k=y/x;
                sum2+=z;
                cb++;
            }
            else
            {
                a[ca].x=x;
                a[ca].y=y;
                a[ca].v=z;
                if(x==0&&y<0) a[ca].k=INF;
                else if(x==0&&y>0) a[ca].k=INF;
                else a[ca].k=y/x;
                sum1+=z;
                ca++;
            }
        }

        ll ans;
        ans=sum1*sum2;
        sort(a,a+ca,cmp);
        sort(b,b+cb,cmp);
        int h=0;
        int l=0;
        while(1)
        {
            if((l==cb||a[h].k>b[l].k)&&h<ca)
               {
                   sum1-=a[h].v;
                   sum2+=a[h].v;
                    h++;
               }
            else if((h==ca||a[h].k<b[l].k)&&l<cb)
              {
                  sum1+=b[l].v;
                  sum2-=b[l].v;
                   l++;
              }
              //cout<<l<<" "<<h<<endl;
              ans=max(ans,sum1*sum2);
              if(l>=cb&&h>=ca) break;
        }
        cout<<ans<<endl;

    }
    return 0;
}