题意:一个强盗要抢劫银行又不想被抓到,所以要进行概率分析求他在不被抓的情况下能抢最多的钱.他给定T(样例个数),N(要抢的银行的个数),P(被抓的概率要小于P)Mj(强盗能抢第j个银行Mj元钱),Pj(强盗抢第j个银行被抓的概率为Pj). 思路:被抓的概率不好直接求出来,但可以直接求出不被抓的概率,则有状态转移方程dp[j] = max(dp[j], dp[j-b[i].money]*b[i].p)表示抢到j元钱被抓的最大的概率是多少.然后逆序遍历第一个小于P的dp的下标就是答案. PS:数组的…

01背包的变形. 先算出硬币面值的总和,然后此题变成求背包容量为V=sum/2时,能装的最多的硬币,然后将剩余的面值和它相减取一个绝对值就是最小的差值. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 50007 ],dp[N]; int…

题意:有N个银行,每抢一个银行,可以获得\(v_i\)的前,但是会有\(p_i\)的概率被抓.现在要把被抓概率控制在\(P\)之下,求最多能抢到多少钱. 分析:0-1背包的变形,把重量变成了概率,因为计算概率需要乘积而非加法,所以不能直接用dp[j]表示概率为j时的最大收益. 令\(dp[i][j]\)表示对前\(i\)个银行,抢到价值为\(j\)还能保持安全的概率,则有递推式: \[dp[i][j] = dp[i-1][j-v[i]]*(1-p[i])\] 第一维其实可以节省下来,因为之和前一…

hdu 1574 RP问题 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1574 分析:01背包的变形. RP可能为负,所以这里分两种情况处理一下就好. 初始化要注意. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define inf 0x3f3f3f3f int…

http://ac.nbutoj.com/Problem/view.xhtml?id=1479 [1479] How many 时间限制: 1000 ms 内存限制: 65535 K 问题描述 There are N numbers, no repeat. All numbers is between 1 and 120, and N is no more than 60. then given a number K(1 <= K <= 100). Your task is to find o…

韩梅梅喜欢满宇宙到处逛街.现在她逛到了一家火星店里,发现这家店有个特别的规矩:你可以用任何星球的硬币付钱,但是绝不找零,当然也不能欠债.韩梅梅手边有104枚来自各个星球的硬币,需要请你帮她盘算一下,是否可能精确凑出要付的款额. 输入格式: 输入第一行给出两个正整数:N(<=104)是硬币的总个数,M(<=102)是韩梅梅要付的款额.第二行给出N枚硬币的正整数面值.数字间以空格分隔. 输出格式: 在一行中输出硬币的面值 V1 <= V2 <= ... <= Vk,满足条件 V1…

Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16522    Accepted Submission(s): 6065 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

The Dwarves of Middle Earth are renowned for their delving and smithy ability, but they are also master builders. During the time of the dragons, the dwarves found that above ground the buildings that were most resistant to attack were truncated squa…

/*Robberies Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13854 Accepted Submission(s): 5111 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the…

题意: 有n<=100双鞋子,分别属于一个牌子,共k<=10个牌子.现有m<=10000钱,问每个牌子至少挑1双,能获得的最大价值是多少? 思路: 分组背包的变形,变成了相反的,每组物品至少挑1件(分组背包问题是至多挑1件). 由于每个牌子至少买1双,那么可以先装一件最便宜的进去,如果有好的再更新(注意每次的容量下限).而且同一双鞋子不能多次购买,这里要用01背包.对于当前容量cap,可能只装了某一牌子的一双鞋子(不一定最便宜),也可能装了多双,也可能只装了那双硬塞进去的最便宜的. 注意…