D - LOOPS Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help her friend Madoka save the world. But because of the plot…

题目大意 有一个人被困在一个 R*C(2<=R,C<=1000) 的迷宫中,起初他在 (1,1) 这个点,迷宫的出口是 (R,C).在迷宫的每一个格子中,他能花费 2 个魔法值开启传送通道.假设他在 (x,y) 这个格子中,开启传送通道之后,有 p_lift[i][j] 的概率被送到 (x,y+1),有 p_down[i][j] 的概率被送到 (x+1,y),有 p_loop[i][j] 的概率被送到 (x,y).问他到出口需要花费的魔法值的期望是多少. 做法分析 令:f[i][j] 表示从…

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…

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…

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…

在拐~ #include <stdio.h> #include <cstring> #include <iostream> #include <map> #include <cmath> template <class T> inline bool rd(T &ret) { char c; int sgn; if(c=getchar(),c==EOF) return 0; while(c!='-'&&(c<…

LOOPS Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1864    Accepted Submission(s): 732 Problem Description Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl). Homura wants to help h…

/ dp求期望的题. 题意:一个软件有s个子系统,会产生n种bug. 某人一天发现一个bug,这个bug属于某种bug,发生在某个子系统中. 求找到所有的n种bug,且每个子系统都找到bug,这样所要的天数的期望. 需要注意的是:bug的数量是无穷大的,所以发现一个bug,出现在某个子系统的概率是1/s, 属于某种类型的概率是1/n. 解法: dp[i][j]表示已经找到i种bug,并存在于j个子系统中,要达到目标状态的天数的期望. 显然,dp[n][s]=0,因为已经达到目标了.而dp[0][…

dp求期望的题. 题意: 有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树, 从结点1出发,开始走,在每个结点i都有3种可能: 1.被杀死,回到结点1处(概率为ki) 2.找到出口,走出迷宫 (概率为ei) 3.和该点相连有m条边,随机走一条 求:走出迷宫所要走的边数的期望值. 设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望.E[1]即为所求. 叶子结点: E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);//因为是到达,…

Collecting Bugs Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers…