Activation HDU - 4089（概率dp）

After 4 years' waiting, the game "Chinese Paladin 5" finally comes out. Tomato is a crazy fan, and luckily he got the first release. Now he is at home, ready to begin his journey. 
But before starting the game, he must first activate the product on the official site. There are too many passionate fans that the activation server cannot deal with all the requests at the same time, so all the players must wait in queue. Each time, the server deals with the request of the first player in the queue, and the result may be one of the following, each has a probability: 
1. Activation failed: This happens with the probability of p1. The queue remains unchanged and the server will try to deal with the same request the next time. 
2. Connection failed: This happens with the probability of p2. Something just happened and the first player in queue lost his connection with the server. The server will then remove his request from the queue. After that, the player will immediately connect to the server again and starts queuing at the tail of the queue.
3. Activation succeeded: This happens with the probability of p3. Congratulations, the player will leave the queue and enjoy the game himself. 
4. Service unavailable: This happens with the probability of p4. Something just happened and the server is down. The website must shutdown the server at once. All the requests that are still in the queue will never be dealt. 
Tomato thinks it sucks if the server is down while he is still waiting in the queue and there are no more than K-1 guys before him. And he wants to know the probability that this ugly thing happens. 
To make it clear, we say three things may happen to Tomato: he succeeded activating the game; the server is down while he is in the queue and there are no more than K-1 guys before him; the server is down while he is in the queue and there are at least K guys before him. 
Now you are to calculate the probability of the second thing.

InputThere are no more than 40 test cases. Each case in one line, contains three integers and four real numbers: N, M (1 <= M <= N <= 2000), K (K >= 1), p1, p2, p3, p4 (0 <= p1, p2, p3, p4 <= 1, p1 + p2 + p3 + p4 = 1), indicating there are N guys in the queue (the positions are numbered from 1 to N), and at the beginning Tomato is at the Mth position, with the probability p1, p2, p3, p4 mentioned above.OutputA real number in one line for each case, the probability that the ugly thing happens. 
The answer should be rounded to 5 digits after the decimal point.

Sample Input

2 2 1 0.1 0.2 0.3 0.4
3 2 1 0.4 0.3 0.2 0.1
4 2 3 0.16 0.16 0.16 0.52

Sample Output

0.30427
0.23280
0.90343
令 dp[i][j] 表示一共有 i 个人，并且 tomato 在第 j 个位置时，达到最终状态的概率。
那么有三种情况：
1.j == 1      : dp[i][j] = p1 * dp[i][j] + p2 * dp[i][i] + p4
2.2 <= j <= k : dp[i][j] = p1 * dp[i][j] + p2 * dp[i][j-1] + p3 * dp[i-1][j-1] + p4
3.k < j <= i  : dp[i][j] = p1 * dp[i][j] + p2 * dp[i][j-1] + p3 * dp[i-1][j-1]
移项化简得
1.j == 1      : dp[i][j] = p21 * dp[i][i] + p41
2.2 <= j <= k : dp[i][j] = p21 * dp[i][j-1] + p31 * dp[i-1][j-1] + p41
3.k < j <= i  : dp[i][j] = p21 * dp[i][j-1] + p31 * dp[i-1][j-1]
其中p21 = p2 / (1 - p1), p31 = p3 / (1 - p1), p41 = p4 / (1 - p1).
此时当计算到 i 时, dp[i-1][] 的值已经全部求出来了，所以可以把后面部分看成常数
1.j == 1      : dp[i][j] = p21 * dp[i][i] + c[j]   c[j] = p41
2.2 <= j <= k : dp[i][j] = p21 * dp[i][j-1] + c[j]   c[j] = p31 * dp[i-1][j-1] + p41
3.k < j <= i  : dp[i][j] = p21 * dp[i][j-1] + c[j]   c[j] = p31 * dp[i-1][j-1]
然后 dp[i][i] 通过 2、3 式推到 dp[i][1]， 然后在用 1 代入 dp[i][1], 求出 dp[i][i], 其式是
dp[i][i] = p21^(i) * dp[i][i] + Σ(p21^(i-x) * c[x])
然后就可以把 dp[i][] 的值都求出来，最后的答案就是dp[n][m]

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 #include <map>
 #include <set>
 #include <list>
 #include <ctime>
 #include <cmath>
 #include <stack>
 #include <queue>
 #include <string>
 #include <vector>
 #include <cstdio>
 #include <bitset>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #define  lowbit(x)  x & (-x)
 #define  mes(a, b)  memset(a, b, sizeof a)
 #define  fi         first
 #define  se         second
 #define  pii        pair<int, int>
 #define  INOPEN     freopen("in.txt", "r", stdin)
 #define  OUTOPEN    freopen("out.txt", "w", stdout)

 typedef unsigned long long int ull;
 typedef long long int ll;
 const int    maxn = 2e3 + ;
 const int    maxm = 1e5 + ;
 const int    mod  = 1e9 + ;
 const ll     INF  = 1e18 + ;
 const int    inf  = 0x3f3f3f3f;
 const double pi   = acos(-1.0);
 const double eps  = 1e-;
 using namespace std;

 int n, m, k;
 int cas, tol, T;

 double c[maxn];
 double pp[maxn];
 double dp[maxn][maxn];

 int main() {
     double p1, p2, p3, p4;
     while(~scanf("%d%d%d%lf%lf%lf%lf", &n, &m, &k, &p1, &p2, &p3, &p4)) {
         if(fabs(p4) <= eps) {
             printf("0.00000\n");
             continue;
         }
         double p21 = p2 / ( - p1);
         double p31 = p3 / ( - p1);
         double p41 = p4 / ( - p1);
         pp[] = 1.0;
         for(int i=; i<=n; i++)
             pp[i] = pp[i-] * p21;
         dp[][] = p41 / ( - p21);
         c[] = p41;
         for(int i=; i<=n; i++) {
             for(int j=; j<=i; j++) {
                 if(j == )        c[j] = p41;
                 else if(j <= k)    c[j] = p31 * dp[i-][j-] + p41;
                 else            c[j] = p31 * dp[i-][j-];
             }
             double tmp = ;
             for(int j=; j<=i; j++) {
                 tmp += pp[i-j] * c[j];
             }
             dp[i][i] = tmp / (1.0 - pp[i]);
             dp[i][] = p21 * dp[i][i] + c[];
             for(int j=; j<i; j++) {
                 dp[i][j] = p21 * dp[i][j-] + c[j];
             }
         }
         for(int i=; i<=n; i++) {
             for(int j=; j<=i; j++) {
                 printf("%f%c", dp[i][j], j==i ? '\n' : ' ');
             }
         }
         printf("%.5f\n", dp[n][m]);
     }
     return ;
 }