An overnight dance in discotheque CodeForces - 814D (几何)

时间：2022-06-01 17:23:34

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 '\n'

#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\n", ans);

}

相关文章