Poj 3744 Scout yyf I (matrix optimization probability), pojyyf

Poj 3744 Scout yyf I (matrix optimization probability), pojyyf

Http://poj.org/problem? Id = 3744

There are n mines. The starting position of a person is 1. The probability of each step is p. The probability of two steps is 1-p, and the position of n mines is given, ask the probability of successful exit from the minefield.

In high school, it should be a simple step-by-step multiplication to calculate the probability. That is, the probability that each mine does not step on is obtained. The last n multiplication is the probability that the mine passes smoothly. Segment the input n locations by 1 ~ Num [1], num [1] + 1 ~ Num [2]... each section has only one thunder num [I], and the probability that each section cannot be thundered is 1-the probability that num [I] Thunder is stepped on.

If dp [I] indicates the probability of stepping on the I Ray, then dp [I] = p * dp [I-1] + (1-p) * dp [I-2], because I is large, it can be optimized by multiplying the matrix.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>//#define LL __int64#define LL long long#define eps 1e-12#define PI acos(-1.0)using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 4010;struct matrix{    double mat[2][2];    void init()    {        mat[0][0] = mat[1][1] = 1;        mat[0][1] = mat[1][0] = 0;    }}a,b,res;int num[15];int n;double p;matrix mul(matrix x, matrix y){    matrix t;    memset(t.mat,0,sizeof(t.mat));    for(int i = 0; i < 2; i++)    {        for(int k = 0; k < 2; k++)        {            if(x.mat[i][k] == 0) continue;            for(int j = 0; j < 2; j++)            {                t.mat[i][j] += x.mat[i][k]*y.mat[k][j];            }        }    }    return t;}matrix pow(matrix a, int n){    matrix t;    t.init();    while(n)    {        if(n&1)            t = mul(t,a);        n >>= 1;        a = mul(a,a);    }    return t;}int main(){    while(~scanf("%d %lf",&n,&p))    {        a.mat[0][0] = p;        a.mat[0][1] = 1-p;        a.mat[1][0] = 1;        a.mat[1][1] = 0;        memset(b.mat,0,sizeof(b.mat));        b.mat[0][0] = 1;        b.mat[1][0] = 0;        for(int i = 1; i <= n; i++)            scanf("%d",&num[i]);        sort(num+1,num+1+n);        double ans;        res = pow(a,num[1]-1);        res = mul(res,b);        ans = 1 - res.mat[0][0];        for(int i = 2; i <= n; i++)        {            res = pow(a,num[i]-num[i-1]-1);            res = mul(res,b);            ans *= (1-res.mat[0][0]);        }        printf("%.7f\n",ans);    }    return 0;}