bzoj1043: [HAOI2008]下落的圆盘

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这道题题目很短，看一下数据范围n<1000，直接枚举圆盘。

对于圆盘i，枚举它上面的圆盘，计算圆盘i被它们覆盖的弧的起始位置与终止位置，然后线段覆盖可得圆盘i被覆盖的周长，就可得答案。

注意圆外离与内含的情况。

#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>#define pi 3.1415926535897932384626433832795using namespace std;struct circle{double r,x,y;}a[10000];struct arc{double a,b;}q[10000];int n,le;double Ans;double di(circle a,circle b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double ang(double R,double d,double r){return acos((R*R+d*d-r*r)/(2*R*d));}bool cmp(arc a,arc b){if (a.a==b.a)return a.b<b.b;else return a.a<b.a;}void add(double l,double r){q[++le].a=l,q[le].b=r;}int main(){freopen("disc.in","r",stdin);freopen("disc.out","w",stdout);scanf("%d",&n);for (int i=1;i<=n;i++)scanf("%lf%lf%lf",&a[i].r,&a[i].x,&a[i].y);for (int i=1;i<=n;i++){bool f=0;le=0;for (int j=i+1;j<=n;j++){double dis=di(a[i],a[j]);if (dis<=a[j].r-a[i].r) {f=1;break;}if (dis>=a[i].r+a[j].r||dis<=a[i].r-a[j].r)continue;double x=a[j].x-a[i].x,y=a[j].y-a[i].y;double a1=atan2(x,y)+pi;double a2=ang(a[i].r,dis,a[j].r);if (a1-a2<0)    add(0,a1+a2),   add(a1-a2+2*pi,2*pi);elseif (a1+a2>2*pi) add(a1-a2,pi*2),add(0,a1+a2-2*pi);else add(a1-a2,a1+a2);}if (f==1)continue;sort(q+1,q+le+1,cmp);double l=-1,r=-1,sum=0;for (int j=1;j<=le;j++)if (q[j].a<=r)r=max(r,q[j].b);else{ sum+=r-l; l=q[j].a; r=q[j].b; }sum+=r-l;Ans+=(2*pi-sum)*a[i].r;}printf("%.3f\n",Ans);return 0;}

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