【Codeforces Round 340 (Div 2)C】【暴力排序枚举】Watering Flowers 2个灌溉器灌溉所有点最小的rr+RR

本文主要是介绍【Codeforces Round 340 (Div 2)C】【暴力排序枚举】Watering Flowers 2个灌溉器灌溉所有点最小的rr+RR，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

A flowerbed has many flowers and two fountains.

You can adjust the water pressure and set any values r₁(r₁ ≥ 0) and r₂(r₂ ≥ 0), giving the distances at which the water is spread from the first and second fountain respectively. You have to set such r₁ and r₂ that all the flowers are watered, that is, for each flower, the distance between the flower and the first fountain doesn't exceed r₁, or the distance to the second fountain doesn't exceed r₂. It's OK if some flowers are watered by both fountains.

You need to decrease the amount of water you need, that is set such r₁ and r₂ that all the flowers are watered and the r₁² + r₂² is minimum possible. Find this minimum value.

Input

The first line of the input contains integers n, x₁, y₁, x₂, y₂ (1 ≤ n ≤ 2000, - 10⁷ ≤ x₁, y₁, x₂, y₂ ≤ 10⁷) — the number of flowers, the coordinates of the first and the second fountain.

Next follow n lines. The i-th of these lines contains integers x_i and y_i ( - 10⁷ ≤ x_i, y_i ≤ 10⁷) — the coordinates of thei-th flower.

It is guaranteed that all n + 2 points in the input are distinct.

Output

Print the minimum possible value r₁² + r₂². Note, that in this problem optimal answer is always integer.

Examples

input

2 -1 0 5 3
0 2
5 2

output

input

output

Note

The first sample is (r₁² = 5, r₂² = 1): $\text{[math]}$ The second sample is (r₁² = 1, r₂² = 32): $\text{[math]}$

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 2020, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
LL X1, Y1, X2, Y2;
LL x[N], y[N];
LL K(LL x) { return x*x; }
LL check(LL r)
{r = r*r;LL R=0;for (int i = 1; i <= n; ++i){if (K(x[i] - X1) + K(y[i] - Y1) <= r);else{gmax(R, K(x[i] - X2) + K(y[i] - Y2));}}return r+R;
}
struct A
{LL x, y, d, D;bool operator < (const A& b)const {return d < b.d;}
}a[N];
int main()
{while (~scanf("%d", &n)){scanf("%lld%lld%lld%lld", &X1,&Y1,&X2,&Y2);for (int i = 1; i <= n; ++i){scanf("%lld%lld", &a[i].x, &a[i].y);a[i].d = K(a[i].x - X1) + K(a[i].y - Y1);a[i].D = K(a[i].x - X2) + K(a[i].y - Y2);}sort(a + 1, a + n + 1); a[0].d = 0;LL ans = 1e18;for (int i = 0; i <= n; ++i){LL R = 0;for (int j = i + 1; j <= n; ++j)gmax(R, a[j].D);gmin(ans, a[i].d + R);}printf("%lld\n", ans);}return 0;
}
/*
【trick&&吐槽】
最近做题太不认真了，这题我竟然没怎么想清楚就写了三分。
写完才发现样例都跑不出。实在是太蠢了！
三分是错误的。
因为在r位于[a[p].d a[p+1].d]范围内时，可能当r为两侧(a[p].d a[p+1].d)的权值时更优。
不满足整体单峰性。【题意】
有2个灌溉器，坐标分别在（X1，Y1）和（X2，Y2）
有n个点需要被灌溉，给出你坐标。
问你，我们如何设置这2个灌溉器的灌溉半径r与R，才能使得所有的点都被灌溉到。
且r^2+R^2尽可能小。【类型】
暴力排序枚举 or 三分【分析】
很显然，r和R都是恰好灌溉到距离其最远的点就好了。
然而，这不意味着r和R是整数。
只意味着r^2与R^2是整数。我们发现点数n很小，只有2000
于是自然想到，我们可以暴力。
如果我们把每个点到1号灌溉源距离的平方设为d，到2号灌溉源距离的平方设为D
那么，我们枚举r^2，只要把所有的d排成升序依次枚举即可。
剩下的灌溉不到的，让2号灌溉源处理。【时间复杂度&&优化】
O(n^2)*/

这篇关于【Codeforces Round 340 (Div 2)C】【暴力排序枚举】Watering Flowers 2个灌溉器灌溉所有点最小的rr+RR的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！