POJ 3744解题报告

Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5466		Accepted: 1510

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

1 0.5
2
2 0.5
2 4

Sample Output

0.5000000
0.2500000

Source

POJ Monthly Contest - 2009.08.23, Simon

           这是一道比较经典的矩阵快速幂优化概率dp的题。给出了n个含有雷的位置。我们不能直接对位置进行概率dp,因为范围比较大，所以可以对n段分别计算，不断叠加。从而求出最终的概率。在这个过程中用矩阵快速幂来优化。该概率dp的状态转移方程比较好写。

       参考代码： 

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<ctime>
#include<cstdlib>
#include<iomanip>
#include<utility>
#define pb push_back
#define mp make_pair
#define CLR(x) memset(x,0,sizeof(x))
#define _CLR(x) memset(x,-1,sizeof(x))
#define REP(i,n) for(int i=0;i<n;i++)
#define Debug(x) cout<<#x<<"="<<x<<" "<<endl
#define REP(i,l,r) for(int i=l;i<=r;i++)
#define rep(i,l,r) for(int i=l;i<r;i++)
#define RREP(i,l,r) for(int i=l;i>=r;i--)
#define rrep(i,l,r) for(int i=1;i>r;i--)
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<11
using namespace std;
int n,a[20],dp[20];
double p;

struct mat
{
    double d[2][2];
} A,B,E;

mat multi(mat a,mat b)
{
    mat ans;
    rep(i,0,2)
    {
        rep(j,0,2)
        {
            ans.d[i][j]=0.0;
            rep(k,0,2)
            if(a.d[i][k]>0&&b.d[k][j]>0)
                ans.d[i][j]+=a.d[i][k]*b.d[k][j];
        }
    }
    return ans;
}

mat quickmulti(mat a,int n)
{
    mat ans=E;
    while(n)
    {
        if(n&1) ans=multi(ans,a);
        a=multi(a,a);
        n>>=1;
    }
    return ans;
}

int main()
{
    E.d[0][0]=E.d[1][1]=1;
    E.d[0][1]=E.d[1][0]=0;
    while(~scanf("%d%lf",&n,&p))
    {
        REP(i,1,n)
        read(a[i]);
        sort(a+1,a+n+1);             //一定要对数组进行排序，因为雷区的位置不是从小到大排列的
        A.d[0][0]=p,A.d[0][1]=1-p;
        A.d[1][0]=1,A.d[1][1]=0;
        B.d[0][0]=1,B.d[0][1]=B.d[1][0]=B.d[1][1]=0;
        REP(i,1,n)
        {
            mat ans;
            if(i==1)
            {
                ans=quickmulti(A,a[i]-1);
                ans=multi(ans,B);
                double p1=ans.d[1][0]*(1-p);
                double p2=p1*p;
                B.d[0][0]=p2,B.d[1][0]=p1;
            }
            else
            {
                if(a[i]-a[i-1]>=2)          //要区分下一个雷区和当前雷区的位置关系，防止出现负数
                {
                    ans=quickmulti(A,a[i]-a[i-1]-2);
                    ans=multi(ans,B);
                    double p1=ans.d[1][0]*(1-p);
                    double p2=p1*p;
                    B.d[0][0]=p2,B.d[1][0]=p1;
                }
                else                    //两个雷区连在一起，则肯定出不去，可以直接break
                {
                    B.d[0][0]=B.d[1][0]=0;
                    break;
                }
            }
        }
        printf("%.7f\n",B.d[1][0]);
    }
}