【概率DP】POJ 3744 Scout YYF I

通道

题意:每次走一步的概率是p，走两步的概率是1-p，然后有n个点上有地雷。问YYF走过雷区的概率是多少

思路： dp[i]=p*dp[i-1]+(1-p)dp[i-2]

转移矩阵： 
  ans[i]    | p ,1-p   |    ans[i-1] 
           =|             | * 
ans[i-1]   | 1 , 0     |   ans[i-2]

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c = getchar() , c == EOF) return false;
    while(c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while(c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return true;
}

const int N = 11;

struct Matrix {
    int r, c;
    double val[N][N];
    void clear() {
        memset(val, 0, sizeof(val));
    }
    void One() {
        for (int i = 0; i < r; ++i)
            val[i][i] = 1;
    }
};
Matrix multi(Matrix a, Matrix b) {
    Matrix re;
    re.r = a.r, re.c = b.c;
    re.clear();
    for (int i = 0; i < re.r; ++i)
    for (int j = 0; j < re.c; ++j)
    for (int k = 0; k < a.c; ++k)
        re.val[i][j] = re.val[i][j] + (a.val[i][k] * b.val[k][j]);
    return re;
}
Matrix Pow(Matrix a, int x) {
    Matrix re;
    re.clear();
    re.r = re.c = a.r;
    re.One();
    while (x > 0) {
        if (x & 1)
            re = multi(re, a);
        a = multi(a, a);
        x >>= 1;
    }
    return re;
}

int n;
int pos[N];
double p;

int main() {
    while (2 == scanf("%d%lf", &n, &p)) {
        for (int i = 0; i < n; ++i) rd(pos[i]);
        sort(pos, pos + n);
        double ans = 1;
        Matrix now;
        now.r = now.c = 2;
        now.val[0][0] = p, now.val[0][1] = 1 - p;
        now.val[1][0] = 1, now.val[1][1] = 0;
        Matrix get = Pow(now, pos[0] - 1);
        ans *= 1 - get.val[0][0];
        for (int i = 1; i < n; ++i) {
            if (pos[i] == pos[i - 1]) continue;
            get = Pow(now, pos[i] - pos[i - 1] - 1);
            ans *= 1 - get.val[0][0];
        }
        printf("%.7f\n", ans);
    }
    return 0;
}