POJ-3744 Scout YYF I 概率DP

　　题目链接：http://poj.org/problem?id=3744

　　简单的概率DP，分段处理，遇到mine特殊处理。f[i]=f[i-1]*p+f[i-2]*(1-p)，i!=w+1，w为mine点。这个概率显然是收敛的，可以转化为(f[i]-f[i-1])/(f[i-1]-f[i-2])=p-1。题目要求精度为1e-7，在分段求的时候我们完全可以控制进度，精度超出了1e-7就不运算下去了。当然此题还可以用矩阵乘法来优化。

　　考虑概率收敛代码：

 //STATUS:C++_AC_0MS_164KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e30;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 double f1,f2,f3;
 double p;
 int n;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,w[],ok;
     while(~scanf("%d%lf",&n,&p)){
         f1=;f2=;
         ok=;
         for(i=;i<n;i++)scanf("%d",&w[i]);
         sort(w,w+n);
         for(i=,j=;j<n;j++){
             if(w[j]==i)ok=;
             for(;i<w[j]- && sign(f2-f1);i++){
                 f3=f2*p+f1*(-p);
                 f1=f2,f2=f3;
             }
             i=w[j]+;
             f2*=-p;
             f1=;
         }
         printf("%.7lf\n",ok?f2:0.0);
     }
     return ;
 }

　　

　　矩阵乘法优化：

 //STATUS:C++_AC_16MS_164KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e30;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 double p;
 int n;

 const int size=;

 struct Matrix{
     double ma[size][size];
     Matrix friend operator * (const Matrix a,const Matrix b){
         Matrix ret;
         mem(ret.ma,);
         int i,j,k;
         for(i=;i<size;i++)
             for(j=;j<size;j++)
                 for(k=;k<size;k++)
                     ret.ma[i][j]=ret.ma[i][j]+a.ma[i][k]*b.ma[k][j];
         return ret;
     }
 }A;

 Matrix mutilpow(int k)
 {
     int i,j;
     Matrix ret;
     mem(ret.ma,);
     for(i=;i<size;i++)
         ret.ma[i][i]=;
     for(;k;k>>=){
         if(k&)ret=ret*A;
         A=A*A;
     }
     return ret;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,w[],ok;
     Matrix S,t;
     double F[];
     while(~scanf("%d%lf",&n,&p)){
         S.ma[][]=,S.ma[][]=;
         S.ma[][]=-p,S.ma[][]=p;
         F[]=,F[]=;
         ok=;
         for(i=;i<n;i++)
             scanf("%d",&w[i]);
         sort(w,w+n);
         for(i=,j=;i<n;i++){
             if(w[i]==j){ok=;break;}
             A=S;
             t=mutilpow(w[i]-j-);
             F[]=(t.ma[][]*F[]+t.ma[][]*F[])*(-p);
             F[]=;
             j=w[i]+;
         }

         printf("%.7lf\n",ok?F[]:0.0);
     }
     return ;
 }