2020 BIT冬训-C++STL I - Watering Flowers CodeForces - 617C

Problem 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.

 

Examples

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

 

 

这题的思路是暴力枚举。先对到一个水池的距离进行排序。因此不在这个水池范围内的花一定在另一个水池内。

由于已经排好序（降序），所以只需要寻找该数据前的数据中与另一个水池中心的最大距离。

需要注意的是不要忽略了所有花都在一个水池而另一个水池一朵花都没有的情况。

这题一开始我是用贪心做的。但是如果出现一朵花在两个水池中点，就会出错。

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef pair<long long,long long> pii;
long long n,dis1,dis2=0x3f3f3f3f,x,y;
pii foun[2];
struct flower{
    long long id,fdis1,fdis2;
}flow[2005];
int cmp(const struct flower & a,const struct flower & b){
    return a.fdis2>b.fdis2;
}
int main(){
    scanf("%lld",&n);
    scanf("%lld%lld%lld%lld",&foun[0].first,&foun[0].second,&foun[1].first,&foun[1].second);
    for(int i=0;i<n;i++){
        scanf("%lld%lld",&x,&y);
        flow[i].fdis1 = (x-foun[0].first)*(x-foun[0].first)+(y-foun[0].second)*(y-foun[0].second);
        flow[i].fdis2 = (x-foun[1].first)*(x-foun[1].first)+(y-foun[1].second)*(y-foun[1].second);
    }
    sort(flow,flow+n,cmp);
    dis2=flow[0].fdis2;
    for(int i=1;i<n;i++){
        dis1=0;
        for(int j=0;j<i;j++){
            if(dis1<flow[j].fdis1){
                dis1=flow[j].fdis1;
            }
        }
        dis2=min(dis2,dis1+flow[i].fdis2);
    }
    if(dis1<flow[n-1].fdis1)
        dis1=flow[n-1].fdis1;
    dis2=min(dis2,dis1);
    printf("%lld",dis2);
}