ZOJ 3329 One Person Game(概率DP)

标签：acm   dp   

题目大意：

There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K₁ faces. Die2 has K₂ faces. Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 to K₁, K₂, K₃ is exactly 1 / K₁, 1 / K₂ and 1 / K₃. You have a counter, and the game is played as follow:

1. Set the counter to 0 at first.
2. Roll the 3 dice simultaneously. If the up-facing number of Die1 is a, the up-facing number of Die2 is b and the up-facing number of Die3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
3. If the counter‘s number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

解题思路：

求概率一般是正推，求期望一般是逆推。这一点可以从题目中体会出来。

设dp[i]表示达到i分时到达目标状态的期望，pk为投掷k分的概率，p0为回到0的概率

则dp[i]=∑(pk*dp[i+k])+dp[0]*p0+1;都和dp[0]有关系，

而且dp[0]就是我们所求，为常数

设dp[i]=A[i]*dp[0]+B[i];

代入上述方程右边得到：dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1 =(∑(pk*A[i+k])+p0)dp[0]+∑(pk*B[i+k])+1;

明显A[i]=(∑(pk*A[i+k])+p0) B[i]=∑(pk*B[i+k])+1

先递推求得A[0]和B[0].

     那么 dp[0]=B[0]/(1-A[0]);

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
const int maxn = 1000 + 10;
double p[maxn];
double A[maxn], B[maxn];
int n, k1, k2, k3, a, b, c;
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        memset(p,0,sizeof(p));
        scanf("%d%d%d%d%d%d%d", &n,&k1,&k2,&k3,&a,&b,&c);
        p[0] = 1.0/k1/k2/k3;
        for(int i=1;i<=k1;i++)
        {
            for(int j=1;j<=k2;j++)
            {
                for(int k=1;k<=k3;k++)
                {
                    if(i != a || j != b || k != c)
                    p[i+j+k] += p[0];
                }
            }
        }
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        for(int i=n;i>=0;i--)
        {
            A[i] = p[0]; B[i] = 1;
            for(int k=1;k<=k1+k2+k3;k++)
            {
                A[i] += p[k] * A[i+k];
                B[i] += p[k] * B[i+k];
            }
        }
        double ans = B[0] / (1 - A[0]);
        printf("%.15lf\n", ans);
    }
    return 0;
}

ZOJ 3329 One Person Game(概率DP)

标签：acm   dp   

原文地址：http://blog.csdn.net/moguxiaozhe/article/details/43486975