POJ 3744 Scout YYF I 機率DP

Scout YYF I

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9564		Accepted: 2799

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 0.522 0.52 4

Sample Output

0.50000000.2500000

Source POJ Monthly Contest - 2009.08.23, Simon

一段一維數軸，在其中n個地方有陷阱。現在一個人從0號位置開始走，走一格的機率為p,走兩格的機率為1-p。求這個人避開所有陷阱的機率。

把路程按照陷阱的先後位置切割成一段一段，把每段避開陷阱的機率乘起來就是答案。

由於n很大，需要用到矩陣快速冪求解。

#include <cstdio>#include <iostream>#include <string.h>#include <string> #include <map>#include <queue>#include <deque>#include <vector>#include <set>#include <algorithm>#include <math.h>#include <cmath>#include <stack>#include <iomanip>#define mem0(a) memset(a,0,sizeof(a))#define meminf(a) memset(a,0x3f,sizeof(a))#define size 2using namespace std;typedef long long ll;typedef long double ld;typedef double db;const int inf=0x3f3f3f3f;  const ll llinf=0x3f3f3f3f3f3f3f3f;   const ld pi=acos(-1.0L);int a[15];struct Matrix {db a[size][size];};Matrix operator*(const Matrix &x,const Matrix &y) {int i,j,k;Matrix ans;for (i=0;i<size;i++) {for (j=0;j<size;j++) {ans.a[i][j]=0.0;for (k=0;k<size;k++) {ans.a[i][j]+=x.a[i][k]*y.a[k][j];//ans.a[i][j]%=m;}}} return ans;}Matrix fastpower(Matrix base,ll index) {Matrix ans,now;int i,j;for (i=0;i<size;i++) {for (j=0;j<size;j++) {if (i==j) ans.a[i][j]=1; else ans.a[i][j]=0;}}if (index<0) return ans;now=base;ll k=index;while (k) {if (k%2) ans=ans*now;now=now*now;k/=2;}return ans;}int main() {int n,i;db p;while (scanf("%d%lf",&n,&p)!=EOF) {Matrix l;for (i=1;i<=n;i++) {scanf("%d",&a[i]);}sort(a+1,a+n+1);a[0]=0;l.a[0][0]=p;l.a[0][1]=1.0-p;l.a[1][0]=1.0;l.a[1][1]=0.0;db tot=1.0;for (i=1;i<=n;i++) {Matrix fp=fastpower(l,a[i]-a[i-1]-1);tot*=1-fp.a[0][0];}printf("%.7lf\n",tot);}return 0;}