An overnight dance in discotheque CodeForces

大意: 给定n个不相交的圆, 求将n个圆划分成两部分, 使得阴影部分面积最大.

贪心, 考虑每个连通块, 最外层大圆分成一部分, 剩余分成一部分一定最优.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+10;
int n, deg[N];
struct {int x,y,r;} a[N];
ll sqr(int x) {return (ll)x*x;}
const double pi = acos(-1);

int main() {
	scanf("%d", &n);
	REP(i,1,n) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].r);
	REP(i,1,n) REP(j,1,i-1) {
		double d = sqrt(sqr(a[i].x-a[j].x)+sqr(a[i].y-a[j].y));
		double dd = abs(a[i].r-a[j].r);
		if (d<=dd) ++deg[a[i].r<a[j].r?i:j];
	}
	double ans = 0;
	REP(i,1,n) {
		if (!deg[i]||(deg[i]&1)) ans+=pi*sqr(a[i].r);
		else ans-=pi*sqr(a[i].r);
	}
	printf("%.12lf
", ans);
}