【HAOI2008】【BZOJ1043】下落的圆盘

1043: [HAOI2008]下落的圆盘

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 887 Solved: 359
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Description

有n个圆盘从天而降，后面落下的可以盖住前面的。求最后形成的封闭区域的周长。看下面这副图, 所有的红色线条的总长度即为所求.

Input

n ri xi y1 … rn xn yn

Output

最后的周长，保留三位小数

Sample Input

2

1 0 0

1 1 0
Sample Output

10.472
HINT

数据规模

n<=1000

Source

暴力枚举每个圆后面的所有圆
对这些圆每个与当前圆计算覆盖的弧度(有两个半径一个圆心距可以用余弦定理)
把算出来的这些弧度存进栈里
最后一起统计一下总共覆盖多少弧度,再计算剩余部分弧长

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define MAXN 1010#define eps 1e-9#define MAXDBL 1e20#define pi acos(-1)using namespace std;int n,top;double ans,sum,temp;struct Bug{    double lb,rb;    bool operator <(const Bug& a)const    {        return lb!=a.lb?lb<a.lb:rb<a.rb;    }}sta[MAXN];struct circle{    double x,y,r;}s[MAXN];inline double sqr(double x){    return x*x;}inline double dis(circle a,circle b){    return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));}inline bool in(circle a,circle b){    return a.r>=b.r+dis(a,b);}inline void circle_cross_circle(circle a,circle b){    double Dis=dis(a,b),tmp,st,bugle;    tmp=(sqr(a.r)-sqr(b.r)+sqr(Dis))/(Dis*2);    st=atan2((a.x-b.x),(a.y-b.y));    bugle=acos(tmp/a.r);    sta[++top]=(Bug){(st-bugle),(st+bugle)};}int main(){    scanf("%d",&n);    for (int i=1;i<=n;i++)  scanf("%lf%lf%lf",&s[i].r,&s[i].x,&s[i].y);    for (int i=1;i<=n;i++)      {        top=0;sum=0;temp=0;        bool flag=0;        for (int j=i+1;j<=n;j++)            if (in(s[j],s[i]))            {                flag=1;                break;            }        if (flag)   continue;        for (int j=i+1;j<=n;j++)            if (!in(s[i],s[j])&&s[i].r+s[j].r>=dis(s[i],s[j]))  circle_cross_circle(s[i],s[j]);        for (int j=1;j<=top;j++)        {            while (sta[j].lb<0) sta[j].lb+=2*pi;            while (sta[j].rb<0) sta[j].rb+=2*pi;            if (sta[j].lb>sta[j].rb)    sta[++top]=(Bug){0,sta[j].rb},sta[j].rb=2*pi;        }        sort(sta+1,sta+top+1);        for (int j=1;j<=top;j++)            if (sta[j].lb>temp) sum+=sta[j].lb-temp,temp=sta[j].rb;            else temp=max(temp,sta[j].rb);        sum+=2*pi-temp;        ans+=s[i].r*sum;    }    printf("%.3f\n",ans);}