HDU4089Activation&概率dp总结

problem

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.
Input
There 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 thdpe probability p1, p2, p3, p4 mentioned above.
Output
A 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

解题思路

概率dp问题，难点在于2处，确立合适的状态转移方程，根据状态转移方程求出递推方程。也就是建立模型与解决问题

本题的状态转移方程比较明显
dp[i][1]=dp[i][1]*p1+dp[i][i]*p2+p4;
dp[i][j]=dp[i][j]*p1+dp[i][j-1]*p2+dp[i-1][j-1]*p3+p4;(j<=k)
dp[i][j]=dp[i][j]*p1+dp[i][j-1]*p2+dp[i-1[[j-1]*p3;(j>k)

根据状态转移方程
可以发现，只要求出dp[i][1]就能求出dpi][j]，而dp[i][1]可以通过dp[i][i]求出
这样解决了问题

大佬的详细解题过程

AC代码

#include
#include
using namespace std;
const int maxn=2005;
const double eps=1e-5;
double dp[maxn][maxn];
double c[maxn];
double p[maxn];
double p1,p2,p3,p4,p21,p31,p41;
int n,m,k;
int main(){
    while(~scanf("%d%d%d%lf%lf%lf%lf",&n,&m,&k,&p1,&p2,&p3,&p4)){
        if(p4<eps){
            cout<<"0.00000\n";
            continue;
        }
        p21=p2/(1-p1);
        p31=p3/(1-p1);
        p41=p4/(1-p1);
        p[0]=1;
        for(int i=1;i<=n;i++){
            p[i]=p21*p[i-1];
        }
        dp[1][1]=p4/(1-p1-p2);
        for(int i=2;i<=n;i++){
            for(int j=1;j<=i;j++){
                if(j<=k){
                    c[j]=dp[i-1][j-1]*p31+p41;
                }
                else{
                    c[j]=dp[i-1][j-1]*p31;
                }
            }
            double temp=0;
            for(int j=1;j<=i;j++){
                temp+=p[i-j]*c[j];
            }
            dp[i][i]=temp/(1-p[i]);
            dp[i][1]=p21*dp[i][i]+p41;
            for(int j=2;j<i;j++){
                dp[i][j]=dp[i][j-1]*p21+c[j];
            }
        }
        printf("%.5lf\n",dp[n][m]);
    }


    return 0;
}

反思

概率dp
向前推期望，
向后推概率
可以用来寻找状态转移方程（使得初始条件比较简单）

找到合适的方程后，应该进行化简，使得其便于计算，可以利用特殊点进行计算，初步进行递推，注意公式推导要正确