The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2166    Accepted Submission(s): 773 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician.…

HDU 3591 The trouble of Xiaoqian(多重背包+全然背包) pid=3591">http://acm.hdu.edu.cn/showproblem.php? pid=3591 题意: 有一个具有n种货币的货币系统, 每种货币的面值为val[i]. 如今小杰手上拿着num[1],num[2],-num[n]个第1种,第2种-第n种货币去买价值为T(T<=20000)的商品, 他给售货员总价值>=T的货币,然后售货员(可能,假设小杰给的钱>T,那肯…

hdu 3591  The trouble of Xiaoqian 题意:xiaoqi要买一个T元的东西,当前的货币有N种,xiaoqi对于每种货币有Ci个:题中定义了最小数量即xiaoqi拿去买东西的钱的张数加上店家找的零钱的张数(店家每种货币有无限多张,且找零是按照最小的数量找零的):问xiaoqi买元东西的最小数量? 多重背包+完全背包: 思路:这个最小数量是拿去买东西的张数和找零的张数之和,其实我们只需要将这两个步骤分开,开算出能买T元东西的前i最少f[i]张,这里涉及到容量问题:容量只…

题目大意 有 \(N\) 种不同面值的硬币,分别给出每种硬币的面值 \(v_i\) 和数量 \(c_i\).同时,售货员每种硬币数量都是无限的,用来找零. 要买价格为 \(T\) 的商品,求在交易中最少使用的硬币的个数(指的是交易中给售货员的硬币个数与找回的硬币个数之和). 个数最多不能超过 \(20000\),如果不能实现,输出 \(-1\):否则输出此次交易中使用的最少的硬币个数. 样例 有 \(3\) 种硬币,面值分别为 \(5, 25 50\),个数分别为 \(5, 2, 1\),要买…

The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1472 Accepted Submission(s): 502 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician. She i…

The trouble of Xiaoqian Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1076    Accepted Submission(s): 355 Problem Description In the country of ALPC , Xiaoqian is a very famous mathematician.…

Cactus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2206    Accepted Submission(s): 1039 Problem Description 1. It is a Strongly Connected graph.2. Each edge of the graph belongs to a circle…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2602 Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 77450    Accepted Submission(s): 32095 Problem Description Many years ago , in…

http://acm.hdu.edu.cn/showproblem.php?pid=3594 题意: 一个有向图,判断是否强连通和每条边只在一个环中. 思路: 仙人掌问题. 用Tarjan算法判断强连通分量的时候,记录每节结点的父节点.当找到一个环后,回溯将该环上的所有结点+1,如果有结点出现2次了,则说明不是仙人掌了. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio…

http://acm.hdu.edu.cn/showproblem.php?pid=3594 判断给定的图是否是强连通的,并且每条边都只属于一个连通分量. 判断强连通只需要判断缩点之后顶点数是否为1即可, 然后在缩点的过程中,如果已经产生环,并且当前结点的父节点还有父节点,则必定有多个环, 最后还要判断每个结点都要只属于一个联通分量,否则不符合要求. #include <iostream> #include <cstdio> #include <cmath> #incl…