Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 18013    Accepted Submission(s): 6653 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10933    Accepted Submission(s): 4049 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

http://acm.hdu.edu.cn/showproblem.php?pid=2955 [题意] 有一个强盗要去几个银行偷盗,他既想多抢点钱,又想尽量不被抓到.已知各个银行 的金钱数和被抓的概率,以及强盗能容忍的最大被抓概率.求他最多能偷到多少钱? [思路] 01背包:每个物品代价是每个银行钱的数目,物品的价值是在该银行不被抓的概率 (1-被抓概率),背包容量是所有银行钱的总和.01背包求dp[i]表示获得i的钱不被抓的最大概率.最后从大到小枚举出 dp[i]>=(1-P)这个i就是答案了…

http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能超过背包的容量1-P. 在这个条件下,让装入背包的物品的总价值,也就是bag[i].[v]的和最大 bag.v是每一件物品的价值,bag.p是每件物品的体积 像上面这样想是行不通的.下面有解释 这道题麻烦的是概率这东西没法用个循环表示出来,根据我以往的经验,指望着把给出的测试数据乘上一百或者一万这种…

Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15649    Accepted Submission(s): 5747 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while,…

直接说题意吧.(什么网友bb了半天题都说不清楚) 给了  P  表示大于这个概率一定被抓住.则P表示被抓住的概率.N表示现在有的银行,pi表示被抓的概率嘛. 然后,就看数学了.肯定不能算被抓的概率啊.因为,如果你第一个就被抓了,后面的概率还用算吗,或则说,这样的计算不符合01 背包的特点,那么我们就想一下,作为一个背包的容量,应该是做加减法,而不是像概率那样做乘除啊.那么我们就用银行的价值作为 物体的体积,而概率(1-被抓的概率=逃跑的概率)作为物品的价值啊.为什么转化为逃跑的概率呢?因为正面肛…

A - Robberies Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2955 Appoint description: Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usu…

题目链接:  HDU 1011 树形背包(DP) Starship Troopers 题意:  地图中有一些房间, 每个房间有一定的bugs和得到brains的可能性值, 一个人带领m支军队从入口(房间1)进入, 只有到达某个房间并且将bugs全部杀死, 才能得到相应该的值. 问最多能获得多少可能性值. PS  1). 一支军队能杀死 20 bugs,  当一支军队发生战争之后就不能再到其它地方去了 2) . 不能走回头路 分析:  [树形背包] 用dp[i][j]表示到达房间 i 的军队数为…

Food Problem Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1243    Accepted Submission(s): 368 Problem Description Few days before a game of orienteering, Bell came to a mathematician to sol…