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：
1 //STATUS:C++_AC_0MS_164KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=200010; 37 const int INF=0x3f3f3f3f; 38 const int MOD=10007,STA=8000010; 39 const LL LNF=1LL<<55; 40 const double EPS=1e-14; 41 const double OO=1e30; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 double f1,f2,f3; 59 double p; 60 int n; 61 62 int main(){ 63 // freopen("in.txt","r",stdin); 64 int i,j,w[12],ok; 65 while(~scanf("%d%lf",&n,&p)){ 66 f1=0;f2=1; 67 ok=1; 68 for(i=0;i<n;i++)scanf("%d",&w[i]); 69 sort(w,w+n); 70 for(i=1,j=0;j<n;j++){ 71 if(w[j]==i)ok=0; 72 for(;i<w[j]-1 && sign(f2-f1);i++){ 73 f3=f2*p+f1*(1-p); 74 f1=f2,f2=f3; 75 } 76 i=w[j]+1; 77 f2*=1-p; 78 f1=0; 79 } 80 printf("%.7lf ",ok?f2:0.0); 81 } 82 return 0; 83 }
矩阵乘法优化：
1 //STATUS:C++_AC_16MS_164KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=200010; 37 const int INF=0x3f3f3f3f; 38 const int MOD=10007,STA=8000010; 39 const LL LNF=1LL<<55; 40 const double EPS=1e-14; 41 const double OO=1e30; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 double p; 59 int n; 60 61 const int size=2; 62 63 struct Matrix{ 64 double ma[size][size]; 65 Matrix friend operator * (const Matrix a,const Matrix b){ 66 Matrix ret; 67 mem(ret.ma,0); 68 int i,j,k; 69 for(i=0;i<size;i++) 70 for(j=0;j<size;j++) 71 for(k=0;k<size;k++) 72 ret.ma[i][j]=ret.ma[i][j]+a.ma[i][k]*b.ma[k][j]; 73 return ret; 74 } 75 }A; 76 77 Matrix mutilpow(int k) 78 { 79 int i,j; 80 Matrix ret; 81 mem(ret.ma,0); 82 for(i=0;i<size;i++) 83 ret.ma[i][i]=1; 84 for(;k;k>>=1){ 85 if(k&1)ret=ret*A; 86 A=A*A; 87 } 88 return ret; 89 } 90 91 int main(){ 92 // freopen("in.txt","r",stdin); 93 int i,j,w[12],ok; 94 Matrix S,t; 95 double F[2]; 96 while(~scanf("%d%lf",&n,&p)){ 97 S.ma[0][0]=0,S.ma[0][1]=1; 98 S.ma[1][0]=1-p,S.ma[1][1]=p; 99 F[0]=0,F[1]=1; 100 ok=1; 101 for(i=0;i<n;i++) 102 scanf("%d",&w[i]); 103 sort(w,w+n); 104 for(i=0,j=1;i<n;i++){ 105 if(w[i]==j){ok=0;break;} 106 A=S; 107 t=mutilpow(w[i]-j-1); 108 F[1]=(t.ma[1][0]*F[0]+t.ma[1][1]*F[1])*(1-p); 109 F[0]=0; 110 j=w[i]+1; 111 } 112 113 printf("%.7lf ",ok?F[1]:0.0); 114 } 115 return 0; 116 }