POJ 3744 Scout YYF I Scout YYF I (递推＋矩阵快速幂)

题意

有n个地方有地雷，初始站在1处，有p的概率往前走1步，1-p的概率往前走2步，求安全通过的概率。

思路

首先我们能想到的是直接递推即dp[i]=p∗dp[i−1]+(1−p)∗dp[i−2]然后矩阵快速幂加速一下，但是这样做中间的地雷很难处理。

因为n很小我们就分段求，对每一个地雷求从前一个地雷到通过的概率，比如求通过第x个地雷的概率就是从dp[x−1]+1走到dp[x]+1的概率，因为只有一步和两步，这个概率直接1−dp[x]就可以了。

矩阵就是经典的斐波那契递推矩阵。

代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define LL long long
#define Lowbit(x) ((x)&(-x))
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1|1
#define MP(a, b) make_pair(a, b)
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int maxn = 1e5 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
int d = 2;
struct Matrix
{
double a[100][100];
Matrix()
{
memset(a, 0, sizeof(a));
}
void init()
{
for (int i = 0; i < d; i++) a[i][i] = 1;
}
Matrix operator * (const Matrix &B)const
{
Matrix C;
for(int i = 0; i < d; i++)
for(int j = 0; j < d; j++)
for(int k = 0; k < d; k++)
C.a[i][j] = C.a[i][j] + a[i][k] * B.a[k][j];
return C;
}
Matrix operator ^ (const int &t)const
{
Matrix res, A = (*this);
res.init();
int p = t;
while (p)
{
if (p & 1) res = res * A;
A = A * A;
p >>= 1;
}
return res;
}
void print()
{
for (int i = 0; i < d; i++)
{
for (int j = 0; j < d; j++)
printf("%-3d", a[i][j]);
printf("\n");
}
}
};

int x[12];

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
double p;
while (scanf("%d%lf", &n, &p) != EOF)
{
x[0] = 0;
for (int i = 1; i <= n; i++)
scanf("%d", &x[i]);
sort(x, x + n + 1);
Matrix A, B;
A.a[0][0] = p, A.a[0][1] = 1 - p;
A.a[1][0] = 1, A.a[1][1] = 0;
B.a[0][0] = 1;
B.a[1][0] = 0;
double ans = 1;
for (int i = 1; i <= n; i++)
{
if (x[i] == x[i-1]) continue;
Matrix temp = A;
temp = temp ^ (x[i] - x[i-1]);
temp = temp * B;
ans *= (1 - temp.a[1][0]);
}
printf("%.7f\n", ans);
}
return 0;
}