DP练习（概率，树状，状压）

http://vjudge.net/contest/view.action?cid=51211#overview

花了好长时间了，终于把这个专题做了绝大部分了

最简单的概率DP求期望，从终点推到起点就是了，注意一个坑就是如果p1=1那么他一旦到达这个点，那么就永远走不出去了

 1 //#pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <map>
 3 #include <set>
 4 #include <stack>
 5 #include <queue>
 6 #include <cmath>
 7 #include <ctime>
 8 #include <vector>
 9 #include <cstdio>
10 #include <cctype>
11 #include <cstring>
12 #include <cstdlib>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 #define INF 1e9
17 #define inf (-((LL)1<<40))
18 #define lson k<<1, L, mid
19 #define rson k<<1|1, mid+1, R
20 #define mem0(a) memset(a,0,sizeof(a))
21 #define mem1(a) memset(a,-1,sizeof(a))
22 #define mem(a, b) memset(a, b, sizeof(a))
23 #define FOPENIN(IN) freopen(IN, "r", stdin)
24 #define FOPENOUT(OUT) freopen(OUT, "w", stdout)
25 
26 //typedef __int64 LL;
27 //typedef long long LL;
28 const int MAXN = 1005;
29 const int MAXM = 100005;
30 const double eps = 1e-13;
31 //const LL MOD = 1000000007;
32 
33 double p1[MAXN][MAXN], p2[MAXN][MAXN], p3[MAXN][MAXN], dp[MAXN][MAXN];
34 
35 int main()
36 {
37     int R, C;
38     while(~scanf("%d %d", &R, &C))
39     {
40         for(int i=1;i<=R;i++)
41             for(int j=1;j<=C;j++)
42                 scanf("%lf%lf%lf", &p1[i][j], &p2[i][j], &p3[i][j]);
43         mem0(dp);
44         for(int i=R;i>=1;i--)
45             for(int j=C;j>=1;j--)
46             {
47                 if(i==R && j==C) continue;
48                 if(fabs(p1[i][j] - 1) < eps) continue;
49                 dp[i][j] = (dp[i][j+1]*p2[i][j] + dp[i+1][j]*p3[i][j] + 2) / (1-p1[i][j]) ;
50             }
51         printf("%.3lf\n", dp[1][1]);
52     }
53     return 0;
54 }

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