LightOj_1079 Just another Robbery

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LightOj_1079 Just another Robbery

题目链接题意: 抢银行(这个背景最爱了), 有n家银行, 每家银行抢劫被抓的概率是p[i],你认为当你被抓的概率低于P的时候是安全的. 问, 你最多能抢劫到多少money. 思路: 抽象成背包问题, 每家银行只有两种选择, 要么抢, 要么不抢. 被抓的概率有点难求, 因为还要考虑之前没有被抓.这里换成求互斥事件:不被抓的概率. 概率为权值, money为重量, 状态转移方程: dp[i] = max(dp[i], dp[i - money[j]] * (1 - p[j])) 最后在dp[mon…

zoj 3511 Cake Robbery(线段树)

problemCode=3511" target="_blank" style="">题目链接:zoj 3511 Cake Robbery 题目大意:就是有一个N边形的蛋糕.切M刀,从中挑选一块边数最多的.保证没有两条边重叠. 解题思路:有多少个顶点即为有多少条边,所以直接依照切刀切掉点的个数排序,然后用线段树维护剩下的还有哪些点. #include <cstdio> #include <cstring> #include &…

Just another Robbery(背包)

1079 - Just another Robbery PDF (English) Statistics Forum Time Limit: 4 second(s) Memory Limit: 32 MB As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He…

UVA 707 - Robbery（内存搜索)

UVA 707 - Robbery 题目链接题意:在一个w * h的图上.t个时刻,然后知道一些信息,每一个时刻没有小偷的矩阵位置,问哪些时刻能够唯一确定小偷位置.和确定小偷是否已经逃走,假设没逃走.可是也没有时刻能够能够确定小偷位置,就是不知到思路:记忆化搜索.dp[x][y][ti]表示在x.y位置.ti时刻时候,小偷是否可能出如今这个位置,1表示有可能.0表示没可能,因为小偷一次最多仅仅能上下左右走一步或者不走.所以去dfs一遍就可以最后推断的时候,假设有一个时刻没有一个1,就表示已…

(概率 01背包) Just another Robbery -- LightOJ -- 1079

http://lightoj.com/volume_showproblem.php?problem=1079 Just another Robbery As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated r…

Codeforces Round #414 A. Bank Robbery

A. Bank Robbery time limit per test 2 seconds memory limit per test 256 megabytes A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes. Oleg the bank client loves money (who doesn't),…

Robbery(记忆化搜索)

Robbery Inspector Robstop is very angry. Last night, a bank has been robbed and the robber has not been caught. And this happened already for the third time this year, even though he did everything in his power to stop the robber: as quickly as possi…