解题报告 之 HDU 4089 Activation

解题报告 之 HDU 4089 Activation

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

 

Source

2011 Asia Beijing Regional Contest

题目大意：有N个人排队以激活游戏，而叔叔站是第M个人。大家依次激活，当前第一个人有p1的概率激活失败，则他还是排在队首等待下一次激活；有p2的概率连接失败，则他到队尾重新排队；有p3激活成功，则他离开队伍；有p4服务器崩溃，则整个过程结束。问当服务器崩溃时，叔叔前面的人少于K个的概率。

分析：一道较难也很有意思的概率DP。用dp[i][j]表示有i个人，叔叔在第j个时，达到目标情况的概率。那么容易写出状态转移方程：

j == 1，            dp[i][j]（dp[i][1]）= p1* dp[i][j]+p2*dp[i][i] （移到最后一个）+ p4

2<=j<=K，       dp[i][j] = p1*dp[i][j] + p2*dp[i][j-1]+p3*dp[i-1][j-1]+p4

K+1<=j<=i,      dp[i][j] = p1*dp[i][j] + p2*dp[i][j-1]+p3*dp[i-1][j-1]

移项后得到：

j == 1，            dp[i][j] = p2/(1-p1)*dp[i][i] + p4/(1-p1)

2<=j<=K，       dp[i][j] = p2/(1-p1)*dp[i][j-1]+p3/(1-p1)*dp[i-1][j-1]+p4/(1-p1)

K+1<=j<=i,      dp[i][j] = p2/(1-p1)*dp[i][j-1]+p3/(1-p1)*dp[i-1][j-1]

令：p21=p2/(1-p1) ， p31=p3/(1-p1)  ， p41=p4/(1-p1) 

j == 1，            dp[i][j] =p21 *dp[i][i] + p41  -->  c[j]=p41

2<=j<=K，       dp[i][j] = p21*dp[i][j-1]+p31*dp[i-1][j-1]+p41  -->  c[j]=dp[i-1][j-1]+p41

K+1<=j<=i,      dp[i][j] = p21*dp[i][j-1]+p31*dp[i-1][j-1]  -->  c[j]=p31*dp[i-1][j-1]

（c[j]的意思是，当我们处理 i 时，对于每个 j 的常数项，或者是常数，或者是i-1行的与 i 行无关）

至于我们大致看出了递推的趋势，但有一个致命的问题，就是dp[i][1]是由dp[i][i]表示的。

所以我们需要先求出dp[i][1]，再进行递推。

这里我们采用方程组的观点，注意dp[i][i]可以用dp[i][j-1]和一些常数表示，而 dp[i][j-1] 又可以用  dp[i][j-2]和常数表示，

……， 一直跌待下去最终可以用dp[i][1]和常数表示dp[i][1]，即可以解出dp[i][1]。

由于每一次迭代都要乘上一个p21，所以常数项的系数可以根据迭代次数算出来。

** 注意不能开二维数组，会爆内存，所以必须要采用滚动数组。

** 需要特判 p4 < 1e-5 的情况，此时直接输出0，不然分母会接近于0而被判为Na值导致超时（这个点我也不是很懂。。。）

上代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;

const int MAXN = 2010;
double dp[3][MAXN];
double pp[MAXN], c[MAXN];
//pp表示p21^p的值，后面迭代求dp[i][i]用；c表示位置为j时的常数项，每一步都要递推更新c。

int main()
{
	int n, m, k;
	double p1, p2, p3, p4;

	while(scanf( "%d%d%d%lf%lf%lf%lf", &n, &m, &k, &p1, &p2, &p3, &p4 ) == 7)
	{
		if(p4<1e-5)
		{
			printf( "0.00000\n" );
			continue;
		}

		double p21 = p2 / (1.0 - p1), p31 = p3 / (1.0 - p1), p41 = p4 / (1.0 - p1);
		pp[0] = 1.0;
		for(int i = 1; i <= n; i++) pp[i] = pp[i - 1] * p21;

		c[1] = p41;
		dp[1][1] = p41 / (1.0 - p21);

		for(int i = 2; i <= n; i++)
		{
			//更新c数组
			for(int j = 2; j <= k; j++) c[j] = p31*dp[1][j - 1] + p41;
			for(int j = k + 1; j <= i; j++) c[j] = p31*dp[1][j - 1];

			//迭代法求dp[i][i]
			double tem = c[1] * pp[i - 1];
			for(int j = 2; j <= k; j++) tem += c[j] * pp[i - j];
			for(int j = k + 1; j <= i; j++) tem += c[j] * pp[i - j];
			dp[2][i] = tem / (1.0 - pp[i]);

			//递推求dp[i][..]
			dp[2][1] = p21*dp[2][i] + c[1];
			for(int j = 2; j <= i; j++) dp[2][j] = p21*dp[2][j - 1] + c[j];

			for(int j = 0; j <= n; j++)
			{
				dp[1][j] = dp[2][j];
			}
		}
		printf( "%.5lf\n", dp[1][m] );
	}
	return 0;
}

就是酱紫的，，，叔叔叔叔靠你辣。陶神陶神靠你辣~

kuangbin 大法好。