CodeForces 617C Watering Flowers

C - Watering Flowers

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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.

Sample Input

Input

2 -1 0 5 3
0 2
5 2

Output

6

Input

4 0 0 5 0
9 4
8 3
-1 0
1 4

Output

33

Hint

The first sample is (r₁² = 5, r₂² = 1):  The second sample is (r₁² = 1, r₂² = 32): 

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题意：给定两个圆的圆心和n个点，找到两个圆的半径r1和r2覆盖所有点。问最小的r1^2 + r2^2。

思路：预处理距离，对于一个点来说，要么被圆1覆盖，要么被圆2覆盖，那么枚举下就好了。时间复杂度O(n^2)。不过是可以优化的，先求出距离d2需要的信息，然后扫一遍d1，时间复杂度O(nlogn)。

O(n^2)枚举每一个花朵在第一个喷泉的距离，第一个喷泉距离比它小的 就不用管了。第一个喷泉距离比它大的，那就一定在第二个喷泉范围里面，求最大值。

枚举第一个距离的每一种情况，求最小值。(其中有可能每一个花朵都在第二个喷泉里，这种情况要考虑到。)

代码：

 #include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<cctype> #include <map> #define max(a,b)(a>b?a:b) #define min(a,b)(a<b?a:b) #define INF 0x3f3f3f3f typedef long long ll;   using namespace std;   struct node {     ll d1,d2; }p[2100];   ll dis(ll x,ll a,ll y,ll b) {     return (a-x)*(a-x)+(b-y)*(b-y); }   int main() {     int n,i,j;     ll ans,x1,y1,x2,y2,x,y;     while(scanf("%d%lld%lld%lld%lld",&n,&x1,&y1,&x2,&y2)!=EOF)     {         for(i=1;i<=n;i++)         {              scanf("%lld%lld",&x,&y);              p[i].d1=dis(x,x1,y,y1);              p[i].d2=dis(x,x2,y,y2);         }           ll Min=1*1e18;           for(i=1;i<=n+1;i++)         {             ll m2=0;             for(j=1;j<=n;j++)             {                 if(p[j].d1>p[i].d1)                    m2=max(m2,p[j].d2);             }             Min=min(Min,m2+p[i].d1);         }         printf("%lld\n",Min);     }     return 0; }