LOOPS 概率dp

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hdu3853 LOOPS(概率dp) 2016-05-26 17:37 89人阅读评论(0) 收藏

LOOPS Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 4636 Accepted Submission(s): 1862 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help…

HDU 3853 LOOPS 概率DP入门

LOOPS Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 8453 Accepted Submission(s): 3397 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h…

hdu 3853 LOOPS 概率DP

简单的概率DP入门题代码如下: #include<iostream> #include<stdio.h> #include<algorithm> #include<iomanip> #include<cmath> #include<cstring> #include<vector> #define ll __int64 #define pi acos(-1.0) #define MAX 1003 using namesp…

题意:迷宫是一个R*C的布局,每个格子中给出停留在原地,往右走一个,往下走一格的概率,起点在(1,1),终点在(R,C),每走一格消耗两点能量,求出最后所需要的能量期望简单概率dp 注意原地不动为1的概率! #include<bits/stdc++.h> using namespace std; #define N 1005 ]; double dp[N][N]; int main() { int n,m; ) { ;i<=n;i++) ;j<=m;j++) ;k<;…

[hdu3853]LOOPS(概率dp)

题意:迷宫是一个R*C的布局,每个格子中给出停留在原地,往右走一个,往下走一格的概率,起点在(1,1),终点在(R,C),每走一格消耗两点能量,求出最后所需要的能量期望. 解题关键:概率dp反向求期望,令$dp[i][j]$表示从该位置走到终点的期望能量,$a$为留在该点的位置,$b$为向下走的概率,$c$为向上走的概率,则 $dp[i][j] = a*dp[i][j] + b*dp[i + 1][j] + c*dp[i][j + 1] + 2$ 移项:$dp[i][j] = (b*dp[i +…

hdu 3853 LOOPS (概率dp 逆推求期望)

题目链接 LOOPS Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)Total Submission(s): 2630 Accepted Submission(s): 1081 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to h…

[转]概率DP总结 by kuangbin

概率类题目一直比较弱,准备把kuangbin大师傅总结的这篇题刷一下! 我把下面的代码换成了自己的代码! 原文地址:http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710606.html *************************************************************** 概率DP主要用于求解期望.概率等题目. 转移方程有时候比较灵活. 一般求概率是正推,求期望是逆推.通过题目可以体会到这点. 首先…