HDU-3853 LOOPS 概率DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3853

　　简单概率DP，转移方程：f[i][j]=f[i][j]*p1+f[i][j+1]*p2+f[i+1][j]*p3+2  —>  f[i][j]=(f[i][j+1]*p2+f[i+1][j]*p3+2)/(1-p1).

 1 //STATUS:C++_AC_2828MS_32180KB

 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=1010;

37 const int INF=0x3f3f3f3f;

38 const int MOD= 1000000007,STA=8000010;

39 const LL LNF=1LL<<55;

40 const double EPS=1e-9;

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 f[N][N],p[N][N][3];

59 int n,m;

60 

61 double dfs(int x,int y)

62 {

63     if(f[x][y]>=0)return f[x][y];

64     if(x==n-1 && y==m-1)return f[x][y]=0;

65     f[x][y]=2/(1-p[x][y][0]);

66     if(p[x][y][1]>0)

67         f[x][y]+=dfs(x,y+1)*p[x][y][1]/(1-p[x][y][0]);

68     if(p[x][y][2]>0)

69         f[x][y]+=dfs(x+1,y)*p[x][y][2]/(1-p[x][y][0]);

70     return f[x][y];

71 }

72 

73 int main(){

74  //   freopen("in.txt","r",stdin);

75     int i,j;

76     while(~scanf("%d%d",&n,&m))

77     {

78         mem(p,0);

79         for(i=0;i<n;i++){

80             for(j=0;j<m;j++){

81                 scanf("%lf%lf%lf",&p[i][j][0],&p[i][j][1],&p[i][j][2]);

82             }

83         }

84 

85         for(i=0;i<n;i++)

86             for(j=0;j<m;j++)

87                 f[i][j]=-1;

88         dfs(0,0);

89 

90         printf("%.3lf\n",f[0][0]);

91     }

92     return 0;

93 }