HDU 4631 Sad Love Story

最新推荐文章于 2019-02-27 16:36:00 发布

原创最新推荐文章于 2019-02-27 16:36:00 发布 · 492 阅读

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本文探讨了暴力枚举算法在多集中应用的实例，包括数据结构、数学计算及编程实现，旨在深入理解复杂问题的求解策略。

multiset+暴力枚举

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef __int64 int64;
typedef long long ll;
#define M 500005
#define N 1000005
#define max_inf 0x7f7f7f7f
#define min_inf 0x80808080
#define mod 1000000007 
#define lc rt<<1
#define rc rt<<1|1

struct Point
{
	ll x , y;
	bool operator < (const Point &a) const
	{
		return x < a.x;
	}
};
ll x[M] , y[M];

int main()
{
	int n , t , i;
	scanf("%d",&t);
	while (t--)
	{
		scanf("%d",&n);
		ll a , b , c;
		scanf("%I64d %I64d %I64d",&a,&b,&c);
		for (x[0] = 0 , i = 1 ; i <= n ; i++)x[i] = (x[i-1]*a+b)%c;
		scanf("%I64d %I64d %I64d",&a,&b,&c);
		for (y[0] = 0 , i = 1 ; i <= n ; i++)y[i] = (y[i-1]*a+b)%c;

		Point p;
		multiset<Point> sq;
		ll Min = (ll)1<<60 , ans = 0;
		p.x = x[1] , p.y = y[1];
		sq.insert(p);
		for (i = 2 ; i <= n ; i++)
		{
			p.x = x[i] , p.y = y[i];
			multiset<Point>::iterator it , It;
			It= sq.lower_bound(p);
			for (it = It ; it != sq.end() ; it++)
			{
				ll xx = it->x-x[i];
				xx *= xx;
				if (xx >= Min)break;
				ll yy = it->y-y[i];
				yy *= yy;
				Min = min(Min,yy+xx);
			}
			for (it = It ; it != sq.begin() ;)
			{ 
				it--;
				ll xx = it->x-x[i];
				xx *= xx;
				if (xx >= Min)break;
				ll yy = it->y-y[i];
				yy *= yy;
				Min = min(Min,yy+xx);
			}

			sq.insert(p);
			ans += Min;
		}

		cout << ans << endl;

	}
	return 0;	
}