Codeforces 617C Watering Flowers 【暴力 数据范围】

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 30 25 2

output

6

input

4 0 0 5 09 48 3-1 01 4

output

33

Note

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

恩，这题注意数据范围。

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>#define ll __int64#define INF 0x3f3f3f3f#define inf 1LL<<63-1#define maxn 2020using namespace std;struct node{    ll dis1,dis2;};node pin[maxn];bool cmp(node a,node b){    return a.dis1<b.dis1;}ll dis(ll x0,ll y0,ll x,ll y){    return ((x-x0)*(x-x0)+(y-y0)*(y-y0));}int main(){    int n;    ll x1,y1,x2,y2,x,y;    while(~scanf("%d",&n))    {        scanf("%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2);        for(int i=1;i<=n;++i)        {            scanf("%I64d%I64d",&x,&y);            pin[i].dis1=dis(x,y,x1,y1);            pin[i].dis2=dis(x,y,x2,y2);        }        pin[0].dis1=0;        pin[0].dis2=0;        sort(pin,pin+n+1,cmp);        ll sum=inf,tem=0;        for(int i=0;i<=n;++i)        {            tem=0;            for(int j=i+1;j<=n;++j)                tem=max(tem,pin[j].dis2);            sum=min(sum,tem+pin[i].dis1);        }        if(n==1)            sum=min(pin[1].dis1,pin[1].dis2);        printf("%I64d\n",sum);    }    return 0;}

1. 该LL使得整数文字类型long long。 所以2LL，是一种2型long long。 如果没有LL，字面只会类型int。 这一点很重要，当你做这样的东西：

1 << 401LL << 40

只用文字1，（假设int为32位，你转移超出了整数类型的大小->未定义的行为）。 同1LL，您将类型设置为long long前手，现在将正确地返回2 ^ 40。
本文标题 ：什么是1LL或2LL在C和C ++？
原文地址 ：CodeGo.net/575292/ （关于1LL的原文）