Codeforces989E. A Trance of Nightfall

Posted 2020-11-18

?????????force   eps   spl   AC   IV   ble   ++   ret   ?????????   

$n leq 200$?????????????????????$q leq 200$????????????????????????$m leq 10000$???????????????$x$???????????????????????????????????????????????????$P$?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????$P$?????????$l$????????????????????????????????????????????????????????????$P$?????????????????????????????????????????????????????????????????????????????????

????????????????????????????????????????????????

?????????????????????????????????????????????????????????$A(x,y)$????????????$f(i,j)$??????$i$?????????$j$????????????$f(i,j)=sum_{k=1}^{n}A(i,k)f(i-1,k)$??????????????????????????????????????????????????????$m-1$???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????$f(m,i)$???????????????????????????????????????????????????????????????$P$???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????P???????????????????????????????????????????????????????????????????????????????????????????????????????????????

??????????????????????????????????????????????????????????????????$qn^3logn$?????????????????????$A^{2^i}$????????????????????????????????????????????????????????????$f$?????????$A^{2^i}$????????????????????????????????????$f$?????????????????????$n^2$?????????????????????$n$???????????????

??????????????????????????????????????????????????????????????????????????????????????????????????????????????????

struct line {
    // ax + by + c = 0
    int a, b, c;

    line() : a(0), b(0), c(0) { }
    line(const point &p, const point &q)
        : a(p.y - q.y)
        , b(q.x - p.x)
        , c(q.y * p.x - q.x * p.y)
    {
        int g = gcd(gcd(a, b), c);
        if (g != 1) { a /= g; b /= g; c /= g; }
        if (a < 0) { a = -a; b = -b; c = -c; }
    }

    inline bool contains(const point &p) {
        return a * p.x + b * p.y + c == 0;
    }

    // For sorting & duplicate removing
    inline bool operator < (const line &other) const {
        return a != other.a ? a < other.a :
            b != other.b ? b < other.b : c < other.c;
    }
    inline bool operator == (const line &other) const {
        return a == other.a && b == other.b && c == other.c;
    }
};

????????????longdouble??????????????????????????????

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdio>
  4 #include<math.h>
  5 //#include<complex>
  6 //#include<set>
  7 //#include<queue>
  8 #include<vector>
  9 #include<algorithm>
 10 #include<stdlib.h>
 11 using namespace std;
 12 
 13 #define double long double
 14 #define LL long long
 15 int qread()
 16 {
 17     char c; int s=0,f=1; while ((c=getchar())<???0??? || c>???9???) (c==???-???) && (f=-1);
 18     do s=s*10+c-???0???; while ((c=getchar())>=???0??? && c<=???9???); return s*f;
 19 }
 20 
 21 //Pay attention to ???-??? , LL and double of qread!!!!
 22 
 23 int n;
 24 #define maxn 40011
 25 #define maxm 40011
 26 #define eps 1e-10
 27 bool equ(double a,double b) {return fabs(a-b)<eps;}
 28 bool neq(double a,double b) {return fabs(a-b)>eps;}
 29 bool le(double a,double b) {return a-b<-eps;}
 30 bool leq(double a,double b) {return a-b<eps;}
 31 bool ge(double a,double b) {return a-b>eps;}
 32 bool geq(double a,double b) {return a-b>-eps;}
 33 
 34 struct Point{int x,y;}p[maxn];
 35 struct Line
 36 {
 37     double k,b;
 38     void make(const Point &A,const Point &B)
 39     {
 40         if (A.x==B.x) {k=1e18; b=A.x; return;}
 41         k=1.0*(A.y-B.y)/(A.x-B.x);
 42         b=A.y-k*A.x;
 43     }
 44     bool operator < (const Line &l)
 45     {
 46         if (neq(l.k,k)) return le(k,l.k);
 47         return le(b,l.b);
 48     }
 49     bool operator == (const Line &l) {return equ(l.k,k) && equ(l.b,b);}
 50     bool operator != (const Line &l) {return !((*this)==l);}
 51 }l[maxm]; int len=0;
 52 vector<int> vl[maxm],vp[maxn];
 53 
 54 struct Mat
 55 {
 56     double a[211][211];
 57     Mat() {memset(a,0,sizeof(a));}
 58     Mat operator * (const Mat &b)
 59     {
 60         Mat ans;
 61         for (int i=1;i<=n;i++)
 62             for (int j=1;j<=n;j++)
 63                 for (int k=1;k<=n;k++)
 64                     ans.a[i][j]+=a[i][k]*b.a[k][j];
 65         return ans;
 66     }
 67 }base[20];
 68 
 69 double f[maxn],g[maxn];
 70 void mul(double *f,Mat b)
 71 {
 72     memset(g,0,sizeof(g));
 73     for (int j=1;j<=n;j++)
 74         for (int k=1;k<=n;k++)
 75             g[j]+=f[k]*b.a[j][k];
 76     for (int j=1;j<=n;j++) f[j]=g[j];
 77 }
 78 
 79 int main()
 80 {
 81     n=qread();
 82     for (int i=1;i<=n;i++) {p[i].x=qread(); p[i].y=qread();}
 83     for (int i=1;i<=n;i++)
 84         for (int j=i+1;j<=n;j++)
 85         {
 86             len++;
 87             l[len].make(p[i],p[j]);
 88         }
 89     sort(l+1,l+1+len);
 90     l[0].k=l[len+1].k=-1e18; l[0].b=l[len+1].b=-1e18;
 91     {
 92         int j=0;
 93         for (int i=1;i<=len;i++) if (l[i]!=l[i-1]) l[++j]=l[i];
 94         len=j;
 95     }
 96     
 97     for (int i=1;i<=n;i++)
 98         for (int j=1;j<=len;j++)
 99             if ((neq(l[j].k,1e18) && equ(l[j].k*p[i].x+l[j].b,p[i].y))
100             || (equ(l[j].k,1e18) && equ(p[i].x,l[j].b)))
101                 vl[j].push_back(i),vp[i].push_back(j);
102     
103     for (int i=1;i<=n;i++)
104         for (int j=0,to=vp[i].size();j<to;j++)
105         {
106             int u=vp[i][j];
107             for (int k=0,too=vl[u].size();k<too;k++)
108             {
109                 int x=vl[u][k];
110                 base[0].a[i][x]+=1.0/(to*too);
111             }
112         }
113     
114     for (int i=1;i<=17;i++) base[i]=base[i-1]*base[i-1];
115     int lq=qread(),x,y;
116     while (lq--)
117     {
118         x=qread(); y=qread(); y--;
119         memset(f,0,sizeof(f)); f[x]=1;
120         for (int i=0;i<=17;i++) if ((y>>i)&1) mul(f,base[i]);
121         double ans=0;
122         for (int i=1;i<=len;i++)
123         {
124             double tmp=0;
125             for (int j=0,to=vl[i].size();j<to;j++) tmp+=f[vl[i][j]];
126             ans=max(ans,tmp/vl[i].size());
127         }
128         cout<<ans<<endl;
129     }
130     return 0;
131 }

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