HDU 6127 Hard challenge (几何)

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

Hard challenge

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 633    Accepted Submission(s): 249

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

221 1 11 -1 131 1 11 -1 10-1 0 100

 

Sample Output

11100

 

Source

2017 Multi-University Training Contest - Team 7

官方解释：

平面直角坐标系上有$n$个整点，第$i$个点有一个点权val_ival​i​​，坐标为(x_i,y_i)(x​i​​,y​i​​)，其中不存在任意两点连成的直线经过原点

整点两两之间连有一条线段，线段的权值为其两端点的权值之积。你需要作一条过原点而不过任意一个给定整点的

直线，使得和这条直线相交的线段的权值和最大。

1\leq n\leq5\times10^4,1\leq val_i\leq10^4,|x_i|,|y_i|\leq10^91≤n≤5×10​4​​,1≤val​i​​≤10​4​​,∣x​i​​∣,∣y​i​​∣≤10​9​​。

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，所以把所有点按极角排个序，然后扫一圈就好了。

感觉这个题目的数据是有点水的，很多情况并没有涉及到就能过。比如，经过原点的一条直线上的两个点应该

必须被一条直线分成两在两侧的，而在一侧充计是不对的。还有就是关于原点的话它的权值是不应该被计入的。

也没有精度问题。再就是LL的问题要注意了。解释一下：比如‘左边’权值和为L，‘右边’权值和为R，（先按着y轴为

直线分的，注意y轴的点关于x轴上下点不会在一个区域内的）当扫描一个点的时候，那在该点与原点连线上的所有

都会变动，对于一个点a而言，如果本身在左边，那么要变到右边的话，L-=a.val，R+=a.val。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <queue>using namespace std;typedef long long LL;struct node{    LL x,y;    LL val;    double k;}point[5*10100];double get_k(LL x,LL y){    if(x==0)return -2*atan(1.0);    return atan((double)y/(double)x);}bool Dir(LL x,LL y){    if(x==0)    {        if(y>0)return true;        else return false;    }    if(x>0)return false;    return true;}bool cmp(node a,node b){    return a.k<b.k;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        LL L=0,R=0,num=0;        for(int i=1;i<=n;i++)        {            scanf("%lld%lld%lld",&point[i].x,&point[i].y,&point[i].val);            //if(point[i].x==point[i].y&&point[i].x==0)point[i].val=0;            point[i].k=get_k(point[i].x,point[i].y);            if(Dir(point[i].x,point[i].y))L+=point[i].val;            else R+=point[i].val;        }        sort(point+1,point+1+n,cmp);        num=L*R;        for(int i=1;i<=n;)        {            double kk=point[i].k;            while(point[i].k==kk)            {                if(point[i].x==0)                {                    if(point[i].y>0)                    {                        L-=point[i].val;                        R+=point[i].val;                    }                    else                    {                        R-=point[i].val;                        L-=point[i].val;                    }                }                else if(point[i].x>0)                {                    R-=point[i].val;                    L+=point[i].val;                }                else                {                    R+=point[i].val;                    L-=point[i].val;                }                i++;            }            num=max(num,L*R);        }        printf("%lld\n",num);    }    return 0;}