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hdu4631（set与二分）

2024-07-03 05:59:14   226人阅读

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4631

Sad Love Story

Time Limit: 40000/20000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1590    Accepted Submission(s): 501


Problem Description
There‘s a really sad story.It could be about love or about money.But love will vanish and money will be corroded.These points will last forever.So this time it is about points on a plane.
We have a plane that has no points at the start.
And at the time i,we add point pi(xi, yi).There is n points in total.
Every time after we add a point,we should output the square of the distance between the closest pair on the plane if there‘s more than one point on the plane.
As there is still some love in the problem setter‘s heart.The data of this problem is randomly generated.
To generate a sequence x1, x2, ..., xn,we let x0 = 0,and give you 3 parameters:A,B,C. Then xi = (xi-1 * A + B) mod C.
The parameters are chosen randomly.
To avoid large output,you simply need output the sum of all answer in one line.
 

Input
The first line contains integer T.denoting the number of the test cases.
Then each T line contains 7 integers:n Ax Bx Cx Ay By Cy.
Ax,Bx,Cx is the given parameters for x1, ..., xn.
Ay,By,Cy is the given parameters for y1, ..., yn.
T <= 10. 
n <= 5 * 105.
104 <= A,B,C <= 106.
 

Output
For each test cases,print the answer in a line.
 

Sample Input
2
5 765934 377744 216263 391530 669701 475509
5 349753 887257 417257 158120 699712 268352
 

Sample Output
8237503125
49959926940

Hint

If there are two points coincide,then the distance between the closest pair is simply 0.
 

Source
2013 Multi-University Training Contest 3
 

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不难，不过如果不进行二分硬算的话，肯定会超。所以我们用multiset来优化，用二分查找，找到插入点的位置p，然后从p向右算，再从p向左算。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
#define LL long long 
using namespace std;
const int MAXN=600000;
LL Ax,Bx,Cx,Ay,By,Cy,ans,n;
class Point
{
public:
	LL x,y;
	bool operator < (const Point &rhs) const 
	{
		return x<rhs.x;
	}
};
Point p[MAXN];
int main()
{
	int T;
	while(cin>>T)
	{
		while(T--)
		{
			cin>>n>>Ax>>Bx>>Cx>>Ay>>By>>Cy;
			p[0].x=0;
			p[0].y=0;
			multiset<Point> S;
			S.clear();
			for(int i=1;i<=n;i++)
			{
				p[i].x=(p[i-1].x*Ax+Bx)%Cx;
				p[i].y=(p[i-1].y*Ay+By)%Cy;
			}
			LL mindis;
			mindis=((LL) 1<<62);
			S.insert(p[1]);
			ans=0;
			for(int i=2;i<=n;i++)
			{
				multiset<Point>::iterator pp=S.lower_bound(p[i]),iter;
				for(iter=pp;iter!=S.begin();)
				{
					iter--;
					Point t=*iter;
					LL a=t.x-p[i].x;
					a*=a;
					if(a>=mindis)
						break;
					LL b=t.y-p[i].y;
					b*=b;
					mindis=min(mindis,a+b);
				}
				for(;pp!=S.end();pp++)
				{
					Point t=*pp;
					LL a=t.x-p[i].x;
					a*=a;
					if(a>=mindis)
						break;
					LL b=t.y-p[i].y;
					b*=b;
					mindis=min(mindis,a+b);
				}
				S.insert(p[i]);
				ans+=mindis;
			}
			cout<<ans<<endl;
		}
	}
	return 0;
}




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