poj1787Charlie's Change(多重背包+记录路径+好题)

【poj1787Charlie's Change(多重背包+记录路径+好题)】的更多相关文章

poj1787Charlie's Change(多重背包+记录路径+好题)

Charlie's Change Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3720 Accepted: 1125 Description Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motore…

（多重背包+记录路径）Charlie's Change （poj 1787）

http://poj.org/problem?id=1787 描述 Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. Yo…

01背包记录路径（例题 L3-001 凑零钱 (30分)）

题意: 就是找出来一个字典序最小的硬币集合,且这个硬币集合里面所有硬币的值的和等于题目中的M 题解: 01背包加一下记录路径,如果1硬币不止一个,那我们也不采用多重背包的方式,把每一个1硬币当成一个独立的单位来进行01背包dp 但是我们知道背包dp的路径可能不止一条,而我们要从中得到字典序最小的序列,我的代码中两次不同的排序会得到最大/小字典序降序 == 最小字典序升序 == 最大字典序 1 #include<iostream> 2 #include<queue> 3 #inc…

poj1417 带权并查集 + 背包 + 记录路径

True Liars Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 2713 Accepted: 868 Description After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exha…

完全背包记录路径poj1787 好题

这题有点多重背包的感觉,但还是用完全背包解决,dp[j]表示凑到j元钱时的最大硬币数,pre[j]是前驱,used[j]是凑到j时第i种硬币的用量 △回溯答案时i-pre[i]就是硬币价值 #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]={,,,},num[],ans[]; ],pre[],used[];//记录回溯状态,记录每个状态用了多少硬币 int main(…

poj 1787 背包+记录路径

http://poj.org/problem?id=1787 Charlie's Change Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 4512 Accepted: 1425 Description Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at co…

牛客网暑期ACM多校训练营（第三场） A PACM Team 01背包记录路径

链接:https://www.nowcoder.com/acm/contest/141/A来源:牛客网 Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in the ACM contest, the most important knowledge is about algorithm, followed by coding(impl…