题目链接: http://poj.org/problem?id=3744 Scout YYF I Time Limit: 1000MSMemory Limit: 65536K 问题描述 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…

F - Scout YYF I Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status 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 d…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5062   Accepted: 1370 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…

Scout YYF I Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5020   Accepted: 1355 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,…

poj 3744 Scout YYF I(递推求期望) 题链 题意:给出n个坑,一个人可能以p的概率一步一步地走,或者以1-p的概率跳过前面一步,问这个人安全通过的概率 解法: 递推式: 对于每个坑,我们可以这么定义一个数组: d[i]代表它安全落在位置i的概率,在这个1到max(a[i])的范围中,只有那些坑是不安全的,答案只需求出所有不掉入坑的概率的连乘即可 矩阵: 由于数字范围巨大,需要对递推式进行矩阵连乘加速 | p 1-p | | d[i] | | d[i+1] | | 1 0 | *…

分段的概率DP+矩阵快速幂                        Scout YYF I Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4180   Accepted: 1076 Description YYF is a couragous scout. Now he is on a dangerous mission which is to penetrate into the enemy's base. Af…

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,…

http://poj.org/problem?id=3744 题意: 现在有个屌丝要穿越一个雷区,雷分布在一条直线上,但是分布的范围很大,现在这个屌丝从1出发,p的概率往前走1步,1-p的概率往前走2步,求最后顺利通过雷区的概率. 思路: 首先很容易能得到一个递推式:$dp[i]=p*dp[i-1]+(1-p)*dp[i-2]$.但是直接递推肯定不行,然后我们发现这个很容易构造出矩阵来,但是这样还是太慢. 接下来讲一下如何优化,对于第i个雷,它的坐标为x[i],那么那顺利通过它的话,只能在x[i…

题目链接 分析&&题意来自 : http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710586.html 题意: 在一条不满地雷的路上,你现在的起点在1处.在N个点处布有地雷,1<=N<=10.地雷点的坐标范围:[1,100000000]. 每次前进p的概率前进一步,1-p的概率前进1-p步.问顺利通过这条路的概率.就是不要走到有地雷的地方. 分析: 设dp[i]表示到达i点的概率,则 初始值 dp[1]=1. 很容易想到转…

题意: 一条路上,给出n地雷的位置,人起始位置在1,向前走一步的概率p,走两步的概率1-p,踩到地雷就死了,求安全通过这条路的概率. 分析: 如果不考虑地雷的情况,dp[i],表示到达i位置的概率,dp[i]=dp[i-1]*p+dp[i-2]*(1-p),要想不踩地雷求出到达地雷位置的概率tmp,1-tmp就是不踩地雷的情况,问题又来了,位置最大是10^9,普通递推超时,想到了用矩阵优化. #include <map> #include <set> #include <li…