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

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.

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58

#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); }