[POJ2151]Check the difficulty of problems (概率dp)

 1 ///#pragma comment(linker, "/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <map>
 7 #include <set>
 8 #include <bitset>
 9 #include <cmath>
10 #include <numeric>
11 #include <iterator>
12 #include <iostream>
13 #include <cstdlib>
14 #include <functional>
15 #include <queue>
16 #include <stack>
17 #include <string>
18 #include <cctype>
19 using namespace std;
20 #define PB push_back
21 #define MP make_pair
22 #define SZ size()
23 #define ST begin()
24 #define ED end()
25 #define CLR clear()
26 #define ZERO(x) memset((x),0,sizeof(x))
27 typedef long long LL;
28 typedef unsigned long long ULL;
29 typedef pair<int,int> PII;
30 const double EPS = 1e-8;
31 
32 const int MAX_T = 1111;
33 const int MAX_M = 33;
34 
35 int M,T,N;
36 double p[MAX_T][MAX_M],f[MAX_T],any[MAX_T],dp[MAX_T][MAX_M][MAX_M];
37 
38 int main(){
39     while( ~scanf("%d%d%d",&M,&T,&N), M!=0&&T!=0&&N!=0 ) {
40         ZERO(dp);
41         ZERO(f);
42         ZERO(any);
43         for(int i=1;i<=T;i++) {
44             for(int j=1;j<=M;j++) {
45                 scanf("%lf",&p[i][j]);
46             }
47         }
48         for(int i=1;i<=T;i++){
49             dp[i][0][0] = 1.0;
50             for(int j=1;j<=M;j++) {
51                 dp[i][j][0] = dp[i][j-1][0]*(1-p[i][j]);
52             }
53             for(int j=1;j<=M;j++){
54                 for(int k=1;k<=j;k++){
55                     dp[i][j][k] = dp[i][j-1][k]*(1-p[i][j]) + dp[i][j-1][k-1]*p[i][j];
56                 }
57             }
58             for(int j=1;j<=M;j++){
59                 for(int k=1;k<=M;k++){
60                     dp[i][j][k] += dp[i][j][k-1];
61                 }
62             }
63         }
64         
65         f[0] = 0.0;
66         any[0] = 1.0;
67         for(int i=1;i<=T;i++){
68             any[i] = any[i-1]*(1.0-dp[i][M][0]);
69         }
70         for(int i=1;i<=T;i++){
71             any[i] = 1.0 - any[i];
72         }
73         any[0] = 0.0;
74         
75         for(int i=1;i<=T;i++){
76             f[i] = f[i-1]*(dp[i][M][M]-dp[i][M][0]) + (1.0-f[i-1]-any[i-1])*(dp[i][M][M]-dp[i][M][N-1]);
77         }
78         
79         printf("%.3f\n",f[T]);
80     }
81     return 0;
82 }