POJ 3744 Scout YYF I

概率$dp$，矩阵优化。

设$dp[i]$为到位置$i$存活的概率，那么如果位置$i$是雷区，$dp[i]=0$，否则$dp[i]=p*dp[i-1]+(1-p)*dp[i-2]$。求出最后一个雷区位置的后一个位置的$dp$值就是答案。长度较大，可以矩阵优化加速一下。输出%$lf$不让过，%$f$过了。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - ‘0‘; c = getchar(); }
}

int n; double p,ans;
int a[20];

struct Matrix
{
    double A[4][4];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= b.C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = c.A[i][j] + A[i][k] * b.A[k][j];
    c.R=R; c.C=b.C;
    return c;
}

void init(double p1,double p2)
{
    Z.A[1][1] = p1, Z.A[1][2] = p2; Z.R = 1; Z.C = 2;
    Y.A[1][1] = 1, Y.A[1][2] = 0, Y.A[2][1] = 0, Y.A[2][2] = 1; Y.R = 2; Y.C = 2;
    X.A[1][1] = 0, X.A[1][2] = 1-p, X.A[2][1] = 1, X.A[2][2] = p; X.R = 2; X.C = 2;
}

void work(int x)
{
    while (x)
    {
        if (x % 2 == 1) Y = Y*X;
        x = x >> 1;
        X = X*X;
    }
    Z = Z*Y;
}

int main()
{
    while(~scanf("%d%lf",&n,&p))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        if(a[1]==1) ans=0;
        else
        {
            bool flag=0;
            for(int i=1;i<n;i++) if(a[i]+1==a[i+1]) flag=1;

            if(flag==1) ans=0;
            else
            {
                double p1=0,p2=1; int pos=1, now=1;
                while(1)
                {
                    if(a[now]!=pos+1)
                    {
                        init(p1,p2);
                        work(a[now]-1-pos);
                        p1=Z.A[1][1]; p2=Z.A[1][2];
                        pos=a[now]-1;
                    }
                    double np1=0,np2=(1-p)*p2;
                    pos=pos+2; p1=np1; p2=np2;
                    now++;
                    if(now==n+1) break;
                }
                ans=p2;
            }
        }
        printf("%.7f\n",ans);
    }
    return 0;
}