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

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

Sample Output

1
1100

Statistic | Submit | Clarifications | Back

题意：平面上一些点，两点连线的价值等于点的价值的乘积，定义经过原点的一条直线的值等于它所相交的线段的价值的总和。

思路：现对这些点进行极角排序，给直线分为两部分，s1和s2分别表示两部分的价值。设排序后的第一个点（分到左边部分即从第一个点右边一点开始）开始逆时针旋转，离开一个点的时候，s1减去那个点的值，s2加上那个点的值；另设一个变量维护当前的扫描始终是180°。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <cmath>

#include <vector>

#include <queue>

#include <set>

#include <map>

using namespace std;

#define ll long long

const double pi=acos(-1.0);

const double eps=1e-8;

int T;

int n;

struct Point{

    int x,y;

    int val;

    double ang;

}point[200000+50];

double angle(Point a){

    double x=abs(a.x);

    double y=abs(a.y);

    if (a.x>0&&a.y>0)   return atan(y/x)*180.0/pi;

    else if (a.x<0&&a.y>0)  return 180.0-atan(y/x)*180.0/pi;

    else if (a.x<0&&a.y<0)  return 180.0+atan(y/x)*180.0/pi;

    else if (a.x>0&&a.y<0)  return 360.0-atan(y/x)*180.0/pi;

    else if (a.x==0){

        if (a.y>0)    return 90;

        else if (a.y<0)   return 270;

        else    return -1;

    }

    else if (a.y==0){

        if (a.x>0)    return 0;

        else if (a.x<0)   return 180;

        else    return -1;

    }

    return -1;

}

bool cmp(Point a,Point b){

    return a.ang<b.ang;

}

int main(){

    // freopen("1.txt","r",stdin);

    scanf("%d",&T);

    while (T--){

        scanf("%d",&n);

        ll sum=0;

        for (int i=1;i<=n;i++){

            scanf("%d %d %d",&point[i].x,&point[i].y,&point[i].val);

            sum+=point[i].val;

            point[i].ang=angle(point[i]);

        }

        sort(point+1,point+n+1,cmp);

        for (int i=n+1;i<=n+n;i++){

            point[i]=point[i-n];

            point[i].ang+=360;

        }

        ll s1=0,s2=0;

        int r;

        for (int i=1;i<=n;i++){

            if (point[i].ang-point[1].ang<180){

                s1+=point[i].val;

            }

            else{

                r=i;

                break;

            }

        }

        s2=sum-s1;

        int l=1;

        ll ans=0;

        while (l<=n){

            ans=max(ans,s1*s2);

            s1-=point[l].val;

            s2+=point[l].val;

            l++;

            while (r<=2*n&&point[r].ang-point[l].ang<180){

                s1+=point[r].val;

                s2-=point[r].val;

                r++;

            }

        }

        printf("%lld\n",ans);

   }

}

巴特西

hdu61272017杭电多校第七场1008Hard challenge

Hard challenge

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