Codeforces Round #340 (Div. 2) 617C Watering Flowers(计算几何)

C. Watering Flowers

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 xi and yi ( - 107 ≤ xi, yi ≤ 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 test(s)

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

Note

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

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


题目链接:点击打开链接

给出n个点坐标, 2个圆点坐标, 求最小的r1, r2平方和, 使得所有点包含在两个圆点组成的圆中.

求出各个点到两圆点坐标距离的平方, 按照距离由大到小排序, 枚举到第二个圆点距离最长, 与到第一个圆点距离和最小.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
ll n, x11, y11, x2, y2, tmp, ans = 1e18;
std::vector<pair<ll, ll> > v;
ll dis1(ll x, ll y)
{
return (x - x11) * (x - x11) + (y - y11) * (y - y11);
}
ll dis2(ll x, ll y)
{
return (x - x2) * (x - x2) + (y - y2) * (y - y2);
}
int main(int argc, char const *argv[])
{
cin >> n >> x11 >> y11 >> x2 >> y2;
for(int i = 0; i < n; ++i) {
ll x, y;
cin >> x >> y;
ll d1 = dis1(x, y), d2 = dis2(x, y);
v.push_back(make_pair(d1, d2));
}
sort(v.rbegin(), v.rend());
for(int i = 0; i < n; ++i) {
ans = min(ans, tmp + v[i].first);
tmp = max(tmp, v[i].second);
}
ans = min(ans, tmp);
cout << ans << endl;
return 0;
}