POJ 3744 概率dp+矩阵
http://poj.org/problem?id=3744
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].
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在一条路（数轴）上有n个地雷，一人从1开始出发， 每走一步，有两种情况：步子为1的概率为p；步子为2的概率为1-p。求能安全通过这条路的概率是多少？
解题思路：这是一道概率dp+矩阵的题，dp[i]=dp[i-1]*p+dp[i-2]*(1-p);这个题的数据范围太大，中间又有地雷间隔，所以是不能直接跑的。我们依据地雷的位置把要求的概率看做n段来求：1~a[0],a[0]+1~a[1],,,,,,,a[n-2]+1~a[n-1].每一段的最后一个数为地雷的位置，这样我们求出每个dp[ a[x] ] 的值，最后把所有的1-a[x] 相乘即为答案。对于每段我们用矩阵来做：
值得一提的是：我们把初始矩阵直接看成:a[1]=1,a[0]=1;这样做起来方便的多
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
struct Matrix
{
double m[2][2];
};
Matrix I={1,0,0,1};
Matrix mult_matrix(Matrix a,Matrix b)
{
Matrix c;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
c.m[i][j]=0;
for(int k=0;k<2;k++)
c.m[i][j]+=a.m[i][k]*b.m[k][j];
}
return c;
}
Matrix quick_mod(Matrix a,int n)
{
Matrix c=I;
while(n)
{
if(n&1)
c=mult_matrix(c,a);
a=mult_matrix(a,a);
n>>=1;
}
return c;
}
int a[15],n;
double p;
int main()
{
while(~scanf("%d%lf",&n,&p))
{
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);//顺序
double ans=1.0;
Matrix A={p,1-p,1,0};//初始化转移矩阵
Matrix B=quick_mod(A,a[0]-1);
ans*=(1-B.m[0][0]);
for(int i=1;i<n;i++)
{
Matrix B=quick_mod(A,a[i]-a[i-1]-1);
ans*=(1-B.m[0][0]);
}
printf("%.7lf\n",ans);
}
return 0;
}