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…

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…

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…

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. Your program will be given numbers and…

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…

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…

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…

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565 记录路径可以用一个二维数组,记录改变时的量.然后从后往前可以推得所有的值. #include <iostream> #include <string> #include <cstring> #include <cstdlib> #incl…

题目链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565 01背包打印路径. #include <cstdio> #include <cstring> using namespace std; ; int n,W; ]; int dp[MAXN]; ][MAXN]; int main() { while(sc…

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(…

http://acm.hdu.edu.cn/showproblem.php?pid=6083 题意: 思路: 01背包+路径记录. 题目有点坑,我一开始逆序枚举菜品,然后一直WA,可能这样的话路径记录会有点问题. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<sstream> #include<vector> #in…

传送门 题意 给出n种物品,抢救第\(i\)种物品花费时间\(t_i\),价值\(p_i\),截止时间\(d_i\) 询问抢救的顺序及物品价值和最大值 分析 按\(d_i\)排序的目的是防止以下情况 4 8 100 1 2 100 不排序只能选择第一个物品 (请仔细思考) 那么排序后做一遍背包,排序后选择顺序必定是递增的,求路径时从n往前找,具体见代码 code #include<bits/stdc++.h> using namespace std; #define ll long long…

#define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;int a[3007],b[3007];int dp[3007],vis[3007],path[3007][3007];int c[3007];int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n,t; cin>>n>>t; fo…

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 motorests. Charlie hates change. That is basically the setup of your next task. Your program will be given…

Charlie's Change Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3176   Accepted: 913 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 motores…

题意:有四种硬币,1分,5分,10分,25分,分别有a,b,c,d种,给出一个n分钱,要求你求出组成n分钱最多需要的硬币数量,并且输出组成它的各种硬币的数量...... 学到的东西:这个题目我是用两种方法做的,一个是完全背包,一个是多重背包.做完这个题目,我对背包的理解可以说上了个层次......还有记录路径的方法,反过来求出各个硬币的数量,都是我以前做的题目没有涉及到的....... 要求出各个硬币有多少种,只需要记录路径,在开一个数组统计p-path[p],为什么可以如此?很容易想到path…

这一道题并不难,我们只需要将dp数组先清空,再给dp[0]=1,之后就按照完全背包的模板写 主要是我们要证明着一种方法不会出现把(1+3+4)(1+4+3)当作两种方法,这一点如果自己写过背包的那个表的话肯定知道是不会出先这种情况的 证明如下:(就拿这一道题的第一个例子来举例) 0  1  2  3  4  5  6  7  8  9  10  11  12 d=1      1  1  1  1   1  1  1  1  1  1    1   1    1 d=5       1  1 …

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…

Charlie's Change Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3792   Accepted: 1144 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…

思路: 完全背包,记录路径. 实现: #include <bits/stdc++.h> using namespace std; const int INF = 0x3f3f3f3f; ] = {, , , }; ], m; ], used[], path[]; int main() { ] >> c[] >> c[] >> c[], m || c[] || c[] || c[] || c[]) { memset(used, , sizeof used);…

Time limit 1000 ms Memory limit 30000 kB 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 motorests. Charlie hates change. That is basically the setup of y…

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. Your program will be given numbers and…

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 motorests. Charlie hates change. That is basically the setup of your next task.  Your program will be give…

E. Fire time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Polycarp is in really serious trouble — his house is on fire! It's time to save the most valuable items. Polycarp estimated that it…

http://poj.org/problem?id=3436 题意:题意很难懂.给出P N.接下来N行代表N个机器,每一行有2*P+1个数字 第一个数代表容量,第2~P+1个数代表输入,第P+2到2*P+1是代表输出 输入有三种情况,0,1,2.输出有0,1两种情况输入0代表不能有这个接口,1代表必须要有这个接口,2代表这个接口可有可无.输出0代表有这个接口,1代表没有这个接口.大概输出就是像插头,输入像插座,只有接口吻合才可以相连. 思路:比较简单的最大流,主要是理解题意很难,把每台机器拆成输…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1026 Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth.…

题意:给了几门学科作业.它们的截止提交期限(天数).它们的需要完成的时间(天数),每项作业在截止日期后每拖延一天扣一学分,算最少扣的学分和其完成顺序. 一开始做的时候,只是听说过状态压缩这个神奇的东西,但事实上我并不会用它,所以白白想了一个晚上没想出来,然后就看了一下题解```再见吧朋友又是新的算法要学了. 状态压缩,实际上就是用二进制的方式,对于每一个要考察的状态用0/1表示其完成与否,这样当从 1 遍历到 111```111 的时候,就可以遍历完所有的状态了,在遍历的过程中利用位运算以及状态…

题意:给了n个家庭作业,然后给了每个家庭作业的完成期限和花费的实践,如果完成时间超过了期限,那么就要扣除分数,然后让你找出一个最优方案使扣除的分数最少,当存在多种方案时,输出字典序最小的那种,因为题意已经说了家庭作业的名字是按照字典序从小到大输入的,所以处理起来就好多了. 分析:此题的关键是如何记录路径,具体看代码实现吧! #include<cstdio> #include<cstring> #include<iostream> #include<algorith…

以前写的题了,现在想整理一下,就挂出来了. 题意比较明确,给一张n*m的地图,从左上角(0, 0)走到右下角(n-1, m-1). 'X'为墙,'.'为路,数字为怪物.墙不能走,路花1s经过,怪物需要花费1s+数字大小的时间. 比较麻烦的是需要记录路径.还要记录是在走路还是在打怪. 因为求最短路,所以可以使用bfs. 因为进过每一个点花费时间不同,所以可以使用优先队列. 因为需要记录路径,所以需要开一个数组,来记录经过节点的父节点.当然,记录方法不止一种. 上代码—— #include <cst…

Minimum Transport Cost Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8860    Accepted Submission(s): 2331 Problem Description These are N cities in Spring country. Between each pair of cities…