hdu 6127 Hard challenge(计算几何)

Hard challenge

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1103    Accepted Submission(s): 457

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

解：乘积的和等于和的乘积，首先分成左右两部分，然后遍历所有的点

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<cmath>#include<string>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N =1e5+10;const int inf = 0x3f3f3f3f;struct node{    LL  v;    double x, y, deg;    bool operator<(const node &A)const    {        return deg>A.deg;    }}p[N], a[N], b[N];int main(){    int t;    scanf("%d", &t);    while(t--)    {        int n;        scanf("%d", &n);        int k1=0, k2=0;        LL s1=0, s2=0;        for(int i=0;i<n;i++)        {            scanf("%lf %lf %lld", &p[i].x,&p[i].y,&p[i].v);            if(p[i].x<=0)            {                if(p[i].x==0)                {                    p[i].deg=inf;                    if(p[i].y<0) p[i].deg=-p[i].deg;                }                else p[i].deg=p[i].y/p[i].x;                a[k1++]=p[i],s1+=p[i].v;            }            else            {                p[i].deg=p[i].y/p[i].x;                b[k2++]=p[i],s2+=p[i].v;            }        }        sort(a,a+k1);        sort(b,b+k2);        LL ans=(s1*s2);        int l=0;        LL s3=s1, s4=s2;        for(int i=0;i<k2;i++)        {            while(l<k1&&a[l].deg>=b[i].deg)            {                s1-=a[l].v,s2+=a[l].v;                ans=max(ans,s1*s2);                l++;            }            s1+=b[i].v,s2-=b[i].v;            ans=max(ans,s1*s2);        }        l=0,s1=s3,s2=s4;        for(int i=0;i<k1;i++)        {            while(l<k2&&b[l].deg>=a[i].deg)            {                s2-=b[l].v,s1+=b[l].v;                ans=max(ans,s1*s2);                l++;            }            s1-=a[i].v,s2+=a[i].v;            ans=max(ans,s1*s2);        }        printf("%lld\n",ans);    }    return 0;}