Codeforcces 617C Watering Flowers【贪心+暴力枚举】

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解决通过调整两个圆的半径来覆盖所有点的问题，使半径平方和最小。

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 the i-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): $\text{[math]}$

题目大意：

给你n个点，和两个圆心的位子，让你用两个圆覆盖住所有的点，使得r1*r1+r2*r2最小。

思路：

1、首先我们做出每个点到圆心1的距离记为dis1，和道圆心2的距离记为dis2。

2、然后我们按照dis1降序排序，排序后，如果我们选择第i个点去用圆1来覆盖，那么其实从i+1-n个点都可以用圆1来覆盖了（因为dis1是降序的），那么从第1个点到第i-1个点就需要用圆2来覆盖，那么我们这个时候我们只需要维护一下从第1个点到第i-1个点的dis2最大值即可，那么当前枚举出来的情况的解就是a【i】.dis1+maxn（这个维护的最大值）

3、按照dis1降序排序之后，我们再按照dis2降序排序，按照上述方式同理维护一个最优解即可。注意数据范围比较大，数据类型和初始化要注意一下。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
#define ll __int64
struct node
{
    ll dis1,dis2;
}a[12121];
ll dis(ll x,ll y,ll x2,ll y2)
{
    ll tmp1=(x-x2)*(x-x2)+(y-y2)*(y-y2);
    return tmp1;
}
int cmp(node a,node b)
{
    if(a.dis2==b.dis2)return a.dis1>b.dis1;
    else return a.dis2>b.dis2;
}
int cmp2(node a,node b)
{
    if(a.dis1==b.dis1)return a.dis2>b.dis2;
    else return a.dis1>b.dis1;
}
int main()
{
    int n;
    ll x1,y1,x2,y2;
    while(~scanf("%d%I64d%I64d%I64d%I64d",&n,&x1,&y1,&x2,&y2))
    {
        for(int i=0;i<n;i++)
        {
            ll x,y;
            scanf("%I64d%I64d",&x,&y);
            a[i].dis1=dis(x,y,x1,y1);
            a[i].dis2=dis(x,y,x2,y2);
        }
        sort(a,a+n,cmp);
        ll ans=100000000000000000;
        for(int i=0;i<n;i++)
        {
            ll tmpmaxn=0;
            for(int j=0;j<i;j++)
            {
                tmpmaxn=max(tmpmaxn,a[j].dis1);
            }
            ans=min(ans,tmpmaxn+a[i].dis2);
        }
        sort(a,a+n,cmp2);
        for(int i=0;i<n;i++)
        {
            ll tmpmaxn=0;
            for(int j=0;j<i;j++)
            {
                tmpmaxn=max(tmpmaxn,a[j].dis2);
            }
            ans=min(ans,tmpmaxn+a[i].dis1);
        }
        printf("%I64d\n",ans);
    }
}