poj 3744 Scout YYF I （可能性DP+矩阵高速功率）

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5062		Accepted: 1370

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

题意：人人从1開始走，p的概率走1步，1-p的概率走2步，求不踩雷的概率。地雷数N<=10，坐标范围在100000000内。

思路：人走到第i位置的概率为dp[i]=dp[i-1]*p+dp[i-2]*(1-p)，因为数据范围较大，可有矩阵高速幂高速求出答案;

构造矩阵     | P ,  1 |   即   | dp[n+1] , dp[n] |  =   | dp[n] , dp[n-1] |    *  | P ,  1 |

            | 1-P , 0 |                                                          | 1-P , 0 |

然后将每一个a[i]+1到a[i+1]看做一段单独处理即可。

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn=100000050;

struct node
{
    double mat[2][2];
}A,B,C;

double dp[maxn],p;
int n,a[15];

void input()
{
    for(int i=0;i<n;i++)  scanf("%d",&a[i]);
    sort(a,a+n);
    A.mat[0][0]=p,A.mat[0][1]=1.0;
    A.mat[1][0]=1.0-p,A.mat[1][1]=0.0;
}

node mul(node p,node q)   //  矩阵相乘
{ 
      node ans;
      for(int i=0;i<2;i++)
           for(int j=0;j<2;j++)
           {
                ans.mat[i][j]=0.0;
                for(int k=0;k<2;k++)
                    ans.mat[i][j]+=p.mat[i][k]*q.mat[k][j];
           }
      return ans;
}

node pow(node w,int num)   //  高速幂
{
    node ret=B,c=w;
    int coun=num;
    while(coun)
    {
        if(coun & 1)  ret=mul(ret,c);
        coun>>=1;
        c=mul(c,c);
    }
    return ret;
}

void solve()
{
    int now=1;
    double f1=0.0,f2=1.0;
    for(int i=0;i<n;i++)
    {
         if(a[i]-now>=1)
         {
              node tmp=pow(A,a[i]-1-now);
              f2=(f2*tmp.mat[0][0]+f1*tmp.mat[1][0])*(1.0-p);
              f1=0.0;
              now=a[i]+1;
         }
         else
         {
             printf("0.0000000\n");
             return ;
         }
    }
    printf("%.7f\n",f2);
}

int main()
{
    B.mat[0][0]=1.0,B.mat[0][1]=0.0;
    B.mat[1][0]=0.0,B.mat[1][1]=1.0;
    while(scanf("%d %lf",&n,&p)!=EOF)
    {
         input();
         solve();
    }
    return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。