Codeforces 814D An overnight dance in discotheque【思维】

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本文探讨了一种算法问题，即如何将舞厅中舞者的活动范围进行最优划分，以实现两个时间段内活动区域的最大化。通过对舞者活动范围的圆进行分析与组合，寻找能够使得两个时段覆盖面积之和最大的分配方案。

The crowdedness of the discotheque would never stop our friends from having fun, but a bit more spaciousness won't hurt, will it?

The discotheque can be seen as an infinite xy-plane, in which there are a total of n dancers. Once someone starts moving around, they will move only inside their own movement range, which is a circular area C_i described by a center (x_i, y_i) and a radius r_i. No two ranges' borders have more than one common point, that is for every pair (i, j) (1 ≤ i < j ≤ n) either ranges C_i and C_j are disjoint, or one of them is a subset of the other. Note that it's possible that two ranges' borders share a single common point, but no two dancers have exactly the same ranges.

Tsukihi, being one of them, defines the spaciousness to be the area covered by an odd number of movement ranges of dancers who are moving. An example is shown below, with shaded regions representing the spaciousness if everyone moves at the same time.

$\text{[math]}$

But no one keeps moving for the whole night after all, so the whole night's time is divided into two halves — before midnight and after midnight. Every dancer moves around in one half, while sitting down with friends in the other. The spaciousness of two halves are calculated separately and their sum should, of course, be as large as possible. The following figure shows an optimal solution to the example above.

$\text{[math]}$

By different plans of who dances in the first half and who does in the other, different sums of spaciousness over two halves are achieved. You are to find the largest achievable value of this sum.

Input

The first line of input contains a positive integer n (1 ≤ n ≤ 1 000) — the number of dancers.

The following n lines each describes a dancer: the i-th line among them contains three space-separated integers x_i, y_i and r_i ( - 10⁶ ≤ x_i, y_i ≤ 10⁶, 1 ≤ r_i ≤ 10⁶), describing a circular movement range centered at (x_i, y_i) with radius r_i.

Output

Output one decimal number — the largest achievable sum of spaciousness over two halves of the night.

The output is considered correct if it has a relative or absolute error of at most 10^- 9. Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if $\text{[math]}$ .

Examples

Input

Output

138.23007676

Input

Output

289.02652413

Note

The first sample corresponds to the illustrations in the legend.

题目大意：

让你将一堆圆圈分成两组，然后每一组取面积并集，问两个组并集的面积最大和是多少。

思路;

以被整个覆盖为界限，偶数分出一组，偶数分出一组。

然后最大面积的圈圈和奇数次的圈圈分在一组，另一部分作为另一组。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
const double PI=acos(-1);
struct node
{
    double x,y,r,s;
    int tot;
}a[1500];
int cmp(node a, node b)
{
    return a.r > b.r;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].r);
            a[i].s=PI*a[i].r*a[i].r;
        }
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                if((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)<(a[i].r+a[j].r)*(a[i].r+a[j].r))
                {
                    a[j].tot++;
                }
            }
        }
        double output=0;
        for(int i=0;i<n;i++)
        {
            if(a[i].tot==0)output+=a[i].s;
            else if(a[i].tot%2==1)output+=a[i].s;
            else if(a[i].tot%2==0)output-=a[i].s;
        }
        printf("%.10lf\n",output);
    }
}