记录编号 42641 评测结果 AAAAAAAAAA
题目名称 [NOIP 2010冲刺六]油滴扩展 最终得分 100
用户昵称 Gravatar临轩听雨ゐ 是否通过 通过
代码语言 C++ 运行时间 0.006 s
提交时间 2012-09-27 19:28:41 内存使用 2.50 MiB
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#include <fstream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
 
int n;
double x[6],y[6],r[6];
double a0,b0,a1,b1,v=0.0;
bool hash[6];
double pi=3.141592653;

void work(int i)
{
    r[i]=1000000.0;
    for (int j=0; j<n; j++)
        if (i!=j && hash[j])
        {
            double nn=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
            r[i]=min(r[i],max(0.0,nn-r[j]));
        }
    r[i]=min(r[i],abs(x[i]-a0));
    r[i]=min(r[i],abs(x[i]-a1));
    r[i]=min(r[i],abs(y[i]-b0));
    r[i]=min(r[i],abs(y[i]-b1));
 }
 
void dfs(int t)
{
    if (t>=n)
    {
        double sum=0.0;
        for (int i=0; i<n; i++)
            sum+=pi*r[i]*r[i];
        v=max(v,sum);
        return;
    }
    for (int i=0; i<n; i++)
        if (!hash[i])
        {
            hash[i]=1;
            work(i);
            dfs(t+1);
            hash[i]=0;
            r[i]=0;
        }
 }
 
int main()
{
    ifstream in("oilbox.in");
    ofstream out("oilbox.out");
    in>>n;
    in>>a0>>b0>>a1>>b1;
    for (int i=0; i<n; i++)
        in>>x[i]>>y[i];
    dfs(0);
    double ans=abs(a0-a1)*abs(b0-b1)-v;
        out<<setiosflags(ios::fixed)<<setprecision(0)<<ans<<endl;
    return 0;
}