BZOJ 1043 圆弧覆盖

为按照极角，转化为线段覆盖问题。

这题有个特别混蛋的trick，就是后掉落的把先掉落的完全覆盖了，我就没考虑这个，无限wa啊！！

 

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  1 #include <iostream>

  2 #include <cstring>

  3 #include <cstdio>

  4 #include <algorithm>

  5 #include <cstdlib>

  6 #include <cmath>

  7 #include <vector>

  8 

  9 #define N 1111

 10 #define PI 3.14159265358979323846

 11 #define EPS 1e-7

 12 

 13 using namespace std;

 14 

 15 struct PO

 16 {

 17     double x,y,r;

 18     void prt() {printf("%lf     %lf\n",x,y);}

 19 }p[N];

 20 

 21 struct V

 22 {

 23     int fg; double f;

 24 };

 25 vector<V> v[N];

 26 

 27 int n;

 28 double sum,ans;

 29 bool vis[N];

 30 

 31 inline int dc(double x)

 32 {

 33     if(x>EPS) return 1;

 34     else if(x<-EPS) return -1;

 35     return 0;

 36 }

 37 

 38 inline bool cmp(const V &a,const V &b)

 39 {

 40     if(a.f==b.f) return a.fg>b.fg;

 41     return a.f<b.f;

 42 }

 43 

 44 inline PO operator -(PO a,PO b)

 45 {

 46     a.x-=b.x; a.y-=b.y;

 47     return a;

 48 }

 49 

 50 inline PO operator +(PO a,PO b)

 51 {

 52     a.x+=b.x; a.y+=b.y;

 53     return a;

 54 }

 55 

 56 inline PO operator *(PO a,double k)

 57 {

 58     a.x*=k; a.y*=k;

 59     return a;

 60 }

 61 

 62 inline PO operator /(PO a,double k)

 63 {

 64     a.x/=k; a.y/=k;

 65     return a;

 66 }

 67 

 68 inline double getlen(const PO &a)

 69 {

 70     return sqrt(a.x*a.x+a.y*a.y);

 71 }

 72 

 73 inline double getdis(const PO &a,const PO &b)

 74 {

 75     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

 76 }

 77 

 78 inline PO rotate(const PO &a,double sss,double ccc)

 79 {

 80     PO rt;

 81     rt.x=a.x*ccc-a.y*sss;

 82     rt.y=a.x*sss+a.y*ccc;

 83     return rt;

 84 }

 85 

 86 inline void read()

 87 {

 88     scanf("%d",&n);

 89     for(int i=1;i<=n;i++)

 90         scanf("%lf%lf%lf",&p[i].r,&p[i].x,&p[i].y);

 91 }

 92 

 93 inline void getcpoint(const PO &a,const PO &b,PO &s,PO &t)//求圆的交点，a的逆时针方向的弧上的点 

 94 {

 95     PO ab=b-a;

 96     double d=getlen(ab);

 97     double ccc=(b.r*b.r-d*d-a.r*a.r)/(-2.0*a.r*d);

 98     double sss=sqrt(1.0-ccc*ccc);

 99     ab=ab/d*a.r;

100     s=a+rotate(ab,-sss,ccc);

101     t=a+rotate(ab,sss,ccc);

102 }

103 

104 inline void go()

105 {

106     PO a,b; V c;

107     double f1,f2;

108     for(int i=1;i<=n;i++)

109         for(int j=i+1;j<=n;j++)

110         {

111             double dis=getdis(p[i],p[j]);

112             if(dis+1e-10>p[i].r+p[j].r || p[j].r+1e-10<p[i].r&&dis-1e-10<p[i].r-p[j].r) continue;

113             if(p[i].r<p[j].r+1e-10&&dis-1e-10<p[j].r-p[i].r) {vis[i]=true; break;}

114             getcpoint(p[i],p[j],a,b);

115             f1=atan2(a.y-p[i].y,a.x-p[i].x);

116             f2=atan2(b.y-p[i].y,b.x-p[i].x);

117             if(f1>f2)

118             {

119                 c.f=f1; c.fg=1; v[i].push_back(c);

120                 c.f=PI; c.fg=-1; v[i].push_back(c);

121                 c.f=-PI; c.fg=1; v[i].push_back(c);

122                 c.f=f2; c.fg=-1; v[i].push_back(c);

123             }

124             else

125             {

126                 c.f=f1; c.fg=1; v[i].push_back(c);

127                 c.f=f2; c.fg=-1; v[i].push_back(c);

128             }

129         }

130     for(int i=1;i<=n;i++)

131     {

132         if(vis[i]) continue;

133         sum+=2*PI*p[i].r;

134         sort(v[i].begin(),v[i].end(),cmp);

135         int tmp=0; double last;

136         for(int j=0;j<v[i].size();j++)

137         {

138             tmp+=v[i][j].fg;

139             if(tmp==1&&v[i][j].fg==1) last=v[i][j].f;

140             else if(tmp==0&&v[i][j].fg==-1) ans+=(v[i][j].f-last)*p[i].r;

141         }

142     }

143     printf("%.3lf\n",sum-ans);

144 }

145 

146 int main()

147 {

148     read(),go();

149     return 0;

150 }