POJ3744 Scout YYF I (矩阵优化的概率DP)

【POJ3744 Scout YYF I (矩阵优化的概率DP)】的更多相关文章

POJ3744 Scout YYF I (矩阵优化的概率DP)

Scout YYF I 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…

Scout YYF I POJ - 3744（概率dp）

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…

poj3744 Scout YYF I[概率dp+矩阵优化]

Scout YYF I Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8598 Accepted: 2521 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,…

[Poj3744]Scout YYF I （概率dp + 矩阵乘法）

Scout YYF I Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9552 Accepted: 2793 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,…

POJ3744 Scout YYF I 概率DP+矩阵快速幂

http://poj.org/problem?id=3744 题意:一条路,起点为1,有概率p走一步,概率1-p跳过一格(不走中间格的走两步),有n个点不能走,问到达终点(即最后一个坏点后)不踩坏点的概率为多少.坏点的坐标范围 [1,100000000] 概率dp的算是入门题…其实写起来和以前的矩阵似乎并没有什么区别呢…状态其实还挺好想的. 坏点sort一下:然后把每一个坏点后一格走到一个坏点前一格的概率乘到答案上,再乘一个1-p跳到坏点后,循环即可. 代码 #include<cstdi…

POJ-3744 Scout YYF I 概率DP

题目链接:http://poj.org/problem?id=3744 简单的概率DP,分段处理,遇到mine特殊处理.f[i]=f[i-1]*p+f[i-2]*(1-p),i!=w+1,w为mine点.这个概率显然是收敛的,可以转化为(f[i]-f[i-1])/(f[i-1]-f[i-2])=p-1.题目要求精度为1e-7,在分段求的时候我们完全可以控制进度,精度超出了1e-7就不运算下去了.当然此题还可以用矩阵乘法来优化. 考虑概率收敛代码: //STATUS:C++_AC_0MS_164K…

POJ-3744 Scout YYF I （矩阵优化概率DP）

题目大意:有n颗地雷分布在一条直线上,有个人的起始位置在1,他每次前进1步的概率为p,前进两步的概率为1-p,问他不碰到地雷的概率. 题目分析:定义状态dp(i)表示到了dp(i)的概率,则状态转移方程为dp(i)=p*dp(i-1)+(1-p)*dp(i-2).要想安全通过x(i)到达x(i)+1只能由x(i)-1走两步,所以,可以将整条直线分成n段,那么从x(i-1)+1安全通过第 i 颗地雷概率为1-p(到达x(i)).坐标之间的距离又很大,所以可以用矩阵二分幂优化. 代码如下: # in…