POJ 3744解題報告

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

           這是一道比較經典的矩陣快速冪優化概率dp的題。給出了n個含有雷的位置。我們不能直接對位置進行概率dp,因爲範圍比較大，所以可以對n段分別計算，不斷疊加。從而求出最終的概率。在這個過程中用矩陣快速冪來優化。該概率dp的狀態轉移方程比較好寫。

       參考代碼： 

#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);             //一定要對數組進行排序，因爲雷區的位置不是從小到大排列的
        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)          //要區分下一個雷區和當前雷區的位置關係，防止出現負數
                {
                    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                    //兩個雷區連在一起，則肯定出不去，可以直接break
                {
                    B.d[0][0]=B.d[1][0]=0;
                    break;
                }
            }
        }
        printf("%.7f\n",B.d[1][0]);
    }
}