HDU-Hard challenge

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于 2017-08-21 17:40:59 发布

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分类专栏： ACM_数学

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本文介绍了一道算法题目，要求通过一条不经过任何给定点的直线将平面上的n个点分成两部分，使得连接这些点的所有线段中被该直线穿过的线段价值总和最大。文章提供了解题思路及实现代码。

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Hard challenge

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1487 Accepted Submission(s): 352

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

Sample Output

1
1100

解析：

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。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define LL long long
const int INF=0x3f3f3f3f;
const double pi=acos(-1.0);
using namespace std;


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;
}