Hdu6127 Hard challenge（2017多校第7场）

平面点集的最大线段价值

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本题涉及在平面上给定多个带有价值的点，通过寻找最优直线以最大化穿过这些点形成的线段的价值总和。文章介绍了问题背景、输入输出格式及样例，并提供了完整的代码实现。

Hard challenge

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 286 Accepted Submission(s): 103

Problem Description

There are

n points on the plane, and the

i th 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

i th 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

Sample Output

  
   1
1100

Source

2017 Multi-University Training Contest - Team 7

————————————————————————————————————

题目的意思是给出n个点，n个点两两之间形成线段价值为端点之积，现在过原点做一条直线，价值为穿过的线段的和，求最大价值

思路：很明显答案就是直线两边点的和的积，我们先按角度对所有点排个序，再搞波前缀和，枚举直线穿过每个点，算出两边的价值，把这个点放到较小的一侧，求最大即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <complex>
#include <vector>
#include <bitset>
#include <stack>
#include <queue>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;
const double pi=acos(-1.0);

int n;
struct node
{
    int x,y;
    LL val;
    double z;
    bool operator<(const node &a)const
    {
        return z<a.z;
    }
}a[50009];
LL sum[50009];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%lld",&a[i].x,&a[i].y,&a[i].val);
            if(a[i].x>0)
            {
                a[i].z=atan(1.0*a[i].y/a[i].x);
                if(a[i].z<0) a[i].z+=2*pi;
            }
            else if(a[i].x==0)
            {
                if(a[i].y>0) a[i].z=pi/2;
                else a[i].z=pi*3/2;
            }
            else
            {
                a[i].z=atan(1.0*a[i].y/a[i].x);
                a[i].z+=pi;
            }
        }
        sort(a+1,a+1+n);
        sum[0]=0;
        for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i].val;
        int k=2;
        LL ma=0;
        for(int i=1;i<=n;i++)
        {
            while(1)
            {
                double p=a[k].z-a[i].z;
                if(p<0) p+=2*pi;
                if(p>pi||k==i) break;
                k=(k+1)%n;
                if(!k) k=n;
            }
            LL sum1,sum2;
            if(k>i) sum1=sum[k-1]-sum[i],sum2=sum[n]-sum1-a[i].val;
            else sum2=sum[i-1]-sum[k-1],sum1=sum[n]-sum2-a[i].val;
            if(sum1<sum2) sum1+=a[i].val;
            else sum2+=a[i].val;
            ma=max(ma,sum1*sum2);
        }
        printf("%lld\n",ma);
    }
    return 0;
}