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
",f[0][0]);
91     }
92     return 0;
93 }