## POJ 3744 Scout YYF I （矩阵优化的概率DP）

Scout YYF I
 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3723 Accepted: 928

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

N个有地雷的点的坐标为 x[1],x[2],x[3]x[N].

1~x[1];
x[1]+1~x[2];
x[2]+1~x[3];



x[N-1]+1~x[N].

/*
POJ 3744

C++  0ms 184K
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;

struct Matrix
{
double mat[2][2];
};
Matrix mul(Matrix a,Matrix b)
{
Matrix ret;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
ret.mat[i][j]=0;
for(int k=0;k<2;k++)
ret.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
}
return ret;
}
Matrix pow_M(Matrix a,int n)
{
Matrix ret;
memset(ret.mat,0,sizeof(ret.mat));
for(int i=0;i<2;i++)ret.mat[i][i]=1;
Matrix temp=a;
while(n)
{
if(n&1)ret=mul(ret,temp);
temp=mul(temp,temp);
n>>=1;
}
return ret;
}

int x[30];
int main()
{
int n;
double p;
while(scanf("%d%lf",&n,&p)!=EOF)//POJ上G++要改为cin输入
{
for(int i=0;i<n;i++)
scanf("%d",&x[i]);
sort(x,x+n);
double ans=1;
Matrix tt;
tt.mat[0][0]=p;
tt.mat[0][1]=1-p;
tt.mat[1][0]=1;
tt.mat[1][1]=0;
Matrix temp;

temp=pow_M(tt,x[0]-1);
ans*=(1-temp.mat[0][0]);

for(int i=1;i<n;i++)
{
if(x[i]==x[i-1])continue;
temp=pow_M(tt,x[i]-x[i-1]-1);
ans*=(1-temp.mat[0][0]);
}
printf("%.7lf\n",ans);//POJ上G++要改为%.7f
}
return 0;
}

posted on 2012-10-02 22:20  kuangbin  阅读(...)  评论(...编辑  收藏

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