$n \leq 200$个平面上的点,$q \leq 200$次询问:重复操作$m \leq 10000$次,到达点$x$的概率最大是多少。操作:一开始选点$P$,不一定要是给定点,可以是平面上任一点。然后,选一条穿过给定点至少两个点且穿过$P$的直线$l$,若有多条,等概率选一条;选中一条后,把$P$点移动到这条直线上的某个初始给定点,若有多个等概率选。
以为是不可做几何题,其实挺好玩的
如果知道两个点之间一次性直接到达的概率$A(x,y)$,那么记$f(i,j)$为走$i$步到点$j$的概率,$f(i,j)=\sum_{k=1}^{n}A(i,k)f(i-1,k)$,是个矩阵乘法,可以快速幂优化。后面$m-1$步可以用矩阵算,那第一步应该怎么选初始点呢?可以发现,固定了终点,用矩阵乘法推出了$f(m,i)$后,每条直线的贡献是固定的,而如果我们选的$P$点在几个直线的交汇处,得到的便是这几条直线贡献的平均值,这不如直接选一条贡献最大的直线(这里的选是说,把点P放在这条线上),而这总是可以做到的。因此输出答案最大的一条直线的答案即可。
有个问题,如果每次询问都要快速幂一次,那不久$qn^3logn$了吗?别担心,$A^{2^i}$是可以预处理的,等会要做乘法的时候只需拿$f$和某些$A^{2^i}$乘就好了,而一个一维数组$f$和矩阵的乘法是$n^2$的,如此少一个$n$,可通过。
通过个鬼。直线运算精度误差坑了我一个半小时。。这里上一个标程无误差版的板子:
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; } };
以及我改longdouble终于通过的辣鸡版本:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }