HDU-Hard challenge
来源:互联网 发布:齐次变换矩阵 含义 编辑:程序博客网 时间:2024/06/03 20:38
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
221 1 11 -1 131 1 11 -1 10-1 0 100
Sample Output
11100
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 longconst 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;}
- HDU 6127 Hard challenge
- hdu 6127 Hard challenge
- hdu--6127--Hard challenge
- HDU 6127 Hard challenge
- HDU 6127 Hard challenge
- hdu 6127 Hard challenge
- HDU 6127 Hard challenge
- hdu 6127 Hard challenge
- HDU-Hard challenge
- HDU 6127 Hard challenge
- HDU 6127 Hard challenge (几何)
- 【HDU 6127 Hard challenge】& 斜率
- HDU 6127 Hard challenge(几何)
- HDU 6127 Hard challenge【几何】
- hdu Hard challenge (几何题)
- HDU 6127 Hard challenge (思维)
- HDU 6127 Hard challenge(几何)
- HDU 6127 Hard challenge【计算几何】
- protobuf Message的序列化和反序列化string类型
- 深入理解回调函数的使用
- BZOJ
- 数据类型及大端和小端
- HDU-2017 多校训练赛6-1002-Mindis
- HDU-Hard challenge
- subplot操作实例python脚本源码
- HDU
- PHP实现SVN管理的一些总结
- 数据结构实验之排序七:选课名单
- 237. Delete Node in a Linked List(链表)
- SSH三大框架的工作原理及流程
- 学生信息管理系统—优化错误篇
- SpannableString与SpannableStringBuilder