poj 3744 Scout YYF I （矩阵乘法+概率与期望DP）

Scout YYF I Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8237   Accepted: 2428 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

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题解：矩阵乘法+概率与期望DP

对于不是地雷的位置，我们用矩阵乘法来优化DP

f[i]=f[i-1]*p+f[i-2]*p

对于是地雷的位置x，

f[x+1]=f[x-1]*(1-p)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,a[20]; double p;
struct data{
	double a[3][3];
}e,c,f;
void clear(data &a)
{
	for (int i=1;i<=2;i++)
	 for (int j=1;j<=2;j++) a.a[i][j]=e.a[i][j];
}
data mul(data a,data b)
{
	data c;
	for (int i=1;i<=2;i++)
	 for (int j=1;j<=2;j++) {
	 	c.a[i][j]=0;
	 	c.a[i][j]+=a.a[i][1]*b.a[1][j];
	 	c.a[i][j]+=a.a[i][2]*b.a[2][j];
	 }
	return c;
}
data quickpow(data num,int x)
{
	data ans; clear(ans);
	data base; base=num;
	while (x) {
		if (x&1) ans=mul(ans,base);
		x>>=1;
		base=mul(base,base);
	}
	return ans;
}
int main()
{
	freopen("a.in","r",stdin);
//	freopen("my.out","w",stdout);
	for (int i=1;i<=2;i++) e.a[i][i]=1;
	while (scanf("%d%lf",&n,&p)!=EOF) {
	   c.a[1][1]=0; c.a[1][2]=(1.0-p);
	   c.a[2][1]=1; c.a[2][2]=p;
	   f.a[1][1]=0; f.a[1][2]=1;
	   for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	   sort(a+1,a+n+1);
	   if (a[1]==1) {
	   	printf("%.7lf\n",0);
	   	continue;
	   }
	   data ans; a[0]=0; bool pd=false;
	   for (int i=1;i<=n;i++) {
	   	  if (a[i-1]+1==a[i]) {
	   	  	   pd=true;
	   	  	   break;
			 }
	   	  data ans=quickpow(c,a[i]-a[i-1]-2);
	   	  ans=mul(f,ans);
	   	  f.a[1][1]=0; f.a[1][2]=ans.a[1][2]*(1.0-p);
	   }
	   if (!pd)  printf("%.7lf\n",f.a[1][2]);
	   else printf("%.7lf\n",0);
	}
}