codeforces 814D An overnight dance in discotheque(思维)

An overnight dance in discotheque

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 Ci described by a center (xi, yi) and a radius ri. No two ranges' borders have more than one common point, that is for every pair (i, j) (1 ≤ i < j ≤ n) either ranges Ci and Cj 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.

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.

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 xi, yi and ri ( - 106 ≤ xi, yi ≤ 106, 1 ≤ ri ≤ 106), describing a circular movement range centered at (xi, yi) with radius ri.

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 .

Examples

input

52 1 60 4 12 -1 31 -2 14 -1 1

output

138.23007676

input

80 0 10 0 20 0 30 0 40 0 50 0 60 0 70 0 8

output

289.02652413

Note

The first sample corresponds to the illustrations in the legend.

题意：给你一个坐标轴，里面有一些圆，在这个圆的覆盖区域中如果有奇数个圆覆盖就加上这块覆盖区域，如果有偶数个圆覆盖就减去这块区域，现在可以将

这个平面一分为2，求最大的区域和；

题解：这道题表面上是求最优解问题，实际上是答案是个定值，仔细想一想，首先从最简单的情况分解问题

如果是这个圆的话没有被别的圆覆盖的话就加上，如果这个圆被一个圆覆盖的话也加上，相当于把他放到第二个平面中，如果放到第一个平面中就是减去，但是被他覆盖的

所有圆的面积加起来都比他小（只考虑比当前层数小的一层，嵌套覆盖的情况按照这个思路考虑），相当于每次都把面积大的那个保留了下来

#include<cstdio>#include<cstdlib>#include<cstring>#include<iostream>#include<cmath>#include <bits/stdc++.h>using namespace std;const int N = 3e5+10;typedef long long LL;struct node{    double x, y, r;}p[N];double dist(int a,int b){    return sqrt((p[a].x-p[b].x)*(p[a].x-p[b].x)+(p[a].y-p[b].y)*(p[a].y-p[b].y));}const double pi=acos(-1.0);int main(){    int n;    scanf("%d", &n);    for(int i=1;i<=n;i++) scanf("%lf %lf %lf",&p[i].x,&p[i].y,&p[i].r);    double ans=0;    for(int i=1;i<=n;i++)    {        int cnt=0;        for(int j=1;j<=n;j++)        {            if(i==j) continue;            double d=dist(i,j);            if(d<p[j].r&&p[j].r>p[i].r) cnt++;        }        if(cnt%2==0&&cnt!=0) ans-=p[i].r*p[i].r;        else ans+=p[i].r*p[i].r;    }    printf("%.10f\n",ans*pi);    return 0;}