POJ 3744 Scout YYF I

最新推荐文章于 2019-09-18 20:25:24 发布

原创最新推荐文章于 2019-09-18 20:25:24 发布 · 864 阅读

0 ·

CC 4.0 BY-SA版权

ACM/概率DP 专栏收录该内容

1 篇文章

订阅专栏

Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4090		Accepted: 1050

Description

YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. After overcoming a series difficulties, YYF is now at the start of enemy's famous "mine road". This is a very long road, on which there are numbers of mines. At first, YYF is at step one. For each step after that, YYF will walk one step with a probability of p, or jump two step with a probality of 1- p. Here is the task, given the place of each mine, please calculate the probality that YYF can go through the "mine road" safely.

Input

The input contains many test cases ended with EOF.
Each test case contains two lines.
The First line of each test case is N (1 ≤ N ≤ 10) and p (0.25 ≤ p ≤ 0.75) seperated by a single blank, standing for the number of mines and the probability to walk one step.
The Second line of each test case is N integer standing for the place of N mines. Each integer is in the range of [1, 100000000].

Output

For each test case, output the probabilty in a single line with the precision to 7 digits after the decimal point.

Sample Input

Sample Output

0.5000000
0.2500000

Source

POJ Monthly Contest - 2009.08.23, Simon

概率DP 根据乘法定理进行分段处理

这样每一段只有一个地雷。我们只要求得通过每一段的概率。乘法原理相乘就是答案。
对于每一段，通过该段的概率等于1-踩到该段终点的地雷的概率。

就比如第一段 1~x[1]. 通过该段其实就相当于是到达x[1]+1点。那么p[x[1]+1]=1-p[x[1]].
但是这个前提是p[1]=1,即起点的概率等于1.对于后面的段我们也是一样的假设，这样就乘起来就是答案了。

对于每一段的概率的求法可以通过矩阵乘法快速求出来。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

double p;
double num[2][2];

void fun(int x)
{
    int i,j,k;
    double cnt[2][2]={p,1.0-p,1.0,0.0},tmp[2][2];
    num[0][0]=num[1][1]=1.0;
    num[0][1]=num[1][0]=0.0;
    while(x>0)
    {
        if(x&1)
        {
            for(i=0;i<2;i++)
                for(j=0;j<2;j++)
                 {
                     tmp[i][j]=num[i][j];
                     num[i][j]=0.0;
                 }

            for(i=0;i<2;i++)
                for(j=0;j<2;j++)
                    for(k=0;k<2;k++)
                        num[i][j]+=tmp[i][k]*cnt[k][j];
        }
        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
            {
                tmp[i][j]=cnt[i][j];
                cnt[i][j]=0.0;
            }

        for(i=0;i<2;i++)
            for(j=0;j<2;j++)
                for(k=0;k<2;k++)
                    cnt[i][j]+=tmp[i][k]*tmp[k][j];
        x>>=1;
    }
    return ;
}

int main()
{
    int N,i,cur,flag;
    int X[100]={-100};
    double d[10];
    double ans;

    while(scanf("%d%lf",&N,&p)==2)
    {
        for(i=1;i<=N;i++)
            scanf("%d",&X[i]);
        sort(X+1,X+1+N);
        if(X[1]==1)
        {
            printf("0.0000000\n");
            continue;
        }
        flag=0;
        for(i=1;i<N;i++)
            if(X[i]+1==X[i+1])
            {
                flag=1;
                printf("0.0000000\n");
                break;
            }
        if(flag)
            continue;
        ans=1.0;
        cur=1;
        for(i=1;i<=N;i++)
        {
            d[1]=1.0;   d[2]=p;
            if(X[i]-cur==1)
            {
                ans*=(1-d[2]);
                cur=X[i]+1;
                continue;
            }
            fun(X[i]-cur-1);
            d[3]=num[0][0]*d[2]+num[0][1]*d[1];
            ans*=(1-d[3]);
            cur=X[i]+1;
        }
        printf("%.7lf\n",ans);
    }
    return 0;
}