http://acm.hdu.edu.cn/showproblem.php?pid=4089
Activation
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1500 Accepted Submission(s):
570
Problem Description
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
the 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
学习：当前面的未知数，而后面也有未知数，看是否有常数项，最后列出一元一次方程。
题意：
有n人都是仙剑5的fans，现在要在官网上激活游戏，n个人排成一个队列（其中主角Tomato最初排名为m），
对于队列中的第一个人，在激活的时候有以下五种情况：
1.激活失败：留在队列中继续等待下一次激活（概率p1）
2.失去连接：激活失败，并且出队列然后排到队列的尾部（概率p2）
3.激活成功：出队列（概率p3）
4.服务器瘫：服务器停止服务了，所有人都无法激活了（概率p4）
求服务器瘫痪并且此时Tomato的排名<=k的概率。
题解：
是一个概率题，分析一下题意后发现和“dp求期望”的题目有点像，因为其中都有一种死循环的可能，
该题中，如果总是发生p1概率的情况那就是死循环了。然后想到一个二维dp：
dp[i][j]表示队列中有i个人，Tomato排在第j个，能发生所求事件的概率。
显然，dp[n][m]即为所求。
j == 1 : dp[i][1] = p1*dp[i][1] + p2*dp[i][i] + p4;
2<=j<=k: dp[i][j] = p1*dp[i][j] + p2*dp[i][j-1] + p3*dp[i-1][j-1] + p4;
j > k : dp[i][j] = p1*dp[i][j] + p2*dp[i][j-1] + p3*dp[i-1][j-1];
化简：
j == 1 : dp[i][1] = p*dp[i][i] + p41;
2<=j<=k: dp[i][j] = p*dp[i][j-1] + p31*dp[i-1][j-1] + p41;
j > k : dp[i][j] = p*dp[i][j-1] + p31*dp[i-1][j-1];
其中：
p = p2 / (1 - p1);
p31 = p3 / (1 - p1);
p41 = p4 / (1 - p1);
现在可以循环 i = 1 -> n 递推求解dp[i]，所以在求dp[i]时，dp[i-1]就相当于常数了，
设dp[i][j]的常数项为c[j]：
j == 1 : dp[i][1] = p*dp[i][i] + c[1];
2<=j<=k: dp[i][j] = p*dp[i][j-1] + c[j];
j > k : dp[i][j] = p*dp[i][j-1] + c[j];
在求dp[i]时，就相当于求“i元1次方程组”：
dp[i][1] = p*dp[i][i] + c[1];
dp[i][2] = p*dp[i][1] + c[2];
dp[i][3] = p*dp[i][2] + c[3];
...
dp[i][i] = p*dp[i][i-1] + c[i];
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const double eps=1e-5;
double dp[2000][2000];
double pp[2000],c[2000];
int main()
{
int n,m,k,i,j;
double p1,p2,p3,p4;
while(~scanf("%d%d%d",&n,&m,&k))
{
memset(dp,0,sizeof(dp));
scanf("%lf%lf%lf%lf",&p1,&p2,&p3,&p4);
if(p4<eps)
{
printf("0.00000\n");
continue;
}
dp[1][1]=p4/(1-(p1+p2));
double p=p2/(1-p1);
double p31=p3/(1-p1);
double p41=p4/(1-p1);
c[1]=p41;
pp[0]=1;
for(i=1;i<=n;i++)
pp[i]=p*pp[i-1];
for(i=2;i<=n;i++)
{
for(j=2;j<=k&&j<=i;j++)
c[j]=p31*dp[i-1][j-1]+p41;
for(j=k+1;j<=n&&j<=i;j++)
c[j]=p31*dp[i-1][j-1];
double temp=c[1]*pp[i-1];
for(j=2;j<=n;j++)
temp+=c[j]*pp[i-j];
dp[i][i]=temp/(1-pp[i]);
dp[i][1]=p*dp[i][i]+c[1];
for(j=2;j<i;j++)
dp[i][j]=p*dp[i][j-1]+c[j];
}
printf("%.5lf\n",dp[n][m]);
}
return 0;
}
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