Buy the souvenirs

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1904    Accepted Submission(s): 711

Problem Description
When the winter holiday comes, a lot of people will have a trip. Generally, there are a lot of souvenirs to sell, and sometimes the travelers will buy some ones with pleasure. Not only can they give the souvenirs to their friends and families as gifts, but also can the souvenirs leave them good recollections. All in all, the prices of souvenirs are not very dear, and the souvenirs are also very lovable and interesting. But the money the people have is under the control. They can’t buy a lot, but only a few. So after they admire all the souvenirs, they decide to buy some ones, and they have many combinations to select, but there are no two ones with the same kind in any combination. Now there is a blank written by the names and prices of the souvenirs, as a top coder all around the world, you should calculate how many selections you have, and any selection owns the most kinds of different souvenirs. For instance:

And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).

 
Input
For the first line, there is a T means the number cases, then T cases follow.
In each case, in the first line there are two integer n and m, n is the number of the souvenirs and m is the money you have. The second line contains n integers; each integer describes a kind of souvenir. 
All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.
 
Output
If you can buy some souvenirs, you should print the result with the same formation as “You have S selection(s) to buy with K kind(s) of souvenirs”, where the K means the most kinds of souvenirs you can buy, and S means the numbers of the combinations you can buy with the K kinds of souvenirs combination. But sometimes you can buy nothing, so you must print the result “Sorry, you can't buy anything.”
 
Sample Input
2
4 7
1 2 3 4

4 0
1 2 3 4

 
Sample Output
You have 2 selection(s) to buy with 3 kind(s) of souvenirs.
Sorry, you can't buy anything.
 
题目大意:
 
一共有 n 个纪念品, 现在你有 m 金币, 告诉你 n 个纪念品的价格, 问你最多可以买多少个纪念品(Max),买最多纪念品有多少个组合(sum)
 
思路:
 
dp[i][k][j] = dp[i-1][k][j] + dp[i-1][k-1][j-a[i]]
dp[i][k][j] 代表从前 i 个纪念品中选 k 个最大价值为 j 的组合数
降维
dp[k][j] = dp[k][j] + dp[k-1][j-a[i]]
dp[k][j] 代表选 k 个最大价值为 j 的组合数
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
using namespace std; const int N = ;
const int INF = 0x3fffffff;
const long long MOD = ;
typedef long long LL;
#define met(a,b) (memset(a,b,sizeof(a))) int a[];
int dp[][N]; /// dp[k][j] 代表选 k 个物品,其中价值为 j 的物品的组合数 int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int i, j, k, n, m, Max=; scanf("%d%d", &n, &m); met(a, );
met(dp, ); for(i=; i<=n; i++)
scanf("%d", &a[i]); dp[][] = ;
for(i=; i<=n; i++)
{
for(k=i; k>=; k--)
{
for(j=a[i]; j<=m; j++)
{ dp[k][j] += dp[k-][j-a[i]];
if(dp[k][j]&&(k>Max)) ///如果 dp[k][j] 有值并且 k>Max 更新Max
Max = k;
}
}
} ///Max 代表从 n 个物品中最多可以选 Max 种物品
///sum 代表有选 Max 个物品的总组合数
int sum = ;
for(i=; i<=m; i++)
sum += dp[Max][i]; if(!Max)
printf("Sorry, you can't buy anything.\n");
else
printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", sum, Max);
}
return ;
}

(01背包)Buy the souvenirs (hdu 2126)的更多相关文章

  1. 【01背包变形】Robberies HDU 2955

    http://acm.hdu.edu.cn/showproblem.php?pid=2955 [题意] 有一个强盗要去几个银行偷盗,他既想多抢点钱,又想尽量不被抓到.已知各个银行 的金钱数和被抓的概率 ...

  2. poj3211Washing Clothes(字符串处理+01背包) hdu1171Big Event in HDU(01背包)

    题目链接: id=3211">poj3211  hdu1171 这个题目比1711难处理的是字符串怎样处理,所以我们要想办法,自然而然就要想到用结构体存储.所以最后将全部的衣服分组,然 ...

  3. HDU-2126 Buy the souvenirs

    数组01背包. http://acm.hdu.edu.cn/showproblem.php?pid=2126 http://blog.csdn.net/crazy_ac/article/details ...

  4. 【hdu2955】 Robberies 01背包

    标签:01背包 hdu2955 http://acm.hdu.edu.cn/showproblem.php?pid=2955 题意:盗贼抢银行,给出n个银行,每个银行有一定的资金和抢劫后被抓的概率,在 ...

  5. HDU 2126 Buy the souvenirs (01背包,输出方案数)

    题意:给出t组数据 每组数据给出n和m,n代表商品个数,m代表你所拥有的钱,然后给出n个商品的价值 问你所能买到的最大件数,和对应的方案数.思路: 如果将物品的价格看做容量,将它的件数1看做价值的话, ...

  6. hdu 2126 Buy the souvenirs 二维01背包方案总数

    Buy the souvenirs Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  7. hdu 2126 Buy the souvenirs(记录总方案数的01背包)

    Buy the souvenirs Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  8. hdu 2126 Buy the souvenirs 买纪念品(01背包,略变形)

    题意: 给出一些纪念品的价格,先算出手上的钱最多能买多少种东西k,然后求手上的钱能买k种东西的方案数.也就是你想要买最多种东西,而最多种又有多少种组合可选择. 思路: 01背包.显然要先算出手上的钱m ...

  9. 【HDU 2126】Buy the souvenirs(01背包)

    When the winter holiday comes, a lot of people will have a trip. Generally, there are a lot of souve ...

随机推荐

  1. 汇编中CMP的作用

    假设现在AX寄存器中的数是0002H,BX寄存器中的数是0003H.执行的指令是:CMP  AX,  BX 执行这条指令时,先做用AX中的数减去BX中的数的减法运算.列出二进制运算式子:      0 ...

  2. 异常处理 day 30

    异常处理 一 错误和异常 二 异常处理 2.1 什么是异常处理? 2.2 为何要进行异常处理? 2.3 如何进行异常处理? 三 什么时候用异常处理 异常和错误 part1:程序中难免出现错误,而错误分 ...

  3. Java并发-ThreadGroup获取所有线程

    一:获取当前项目所有线程 public Thread[] findAllThread(){ ThreadGroup currentGroup =Thread.currentThread().getTh ...

  4. Window10系统的安装

    关于系统的安装网上有许多的教程,本文的教程并没有什么特别的.只是将自己在安装过程中遇到的问题记录下来,方便以后观看. 1.下载系统镜像 首先从MSDN上下载windows10镜像.在操作系统Windo ...

  5. Scrapy框架学习笔记

    1.Scrapy简介 Scrapy是用纯Python实现一个为了爬取网站数据.提取结构性数据而编写的应用框架,用途非常广泛. 框架的力量,用户只需要定制开发几个模块就可以轻松的实现一个爬虫,用来抓取网 ...

  6. Vsphere初试——架设Panabit行为管理

    Panabit是目前国内X86平台单板处理能力最高(双向40G).提供免费版本(软件形态),是以DPI为核心优势并发展起来的最专业.上线效果最好.性价比最高的新一代应用网关.Panabit流控引擎,基 ...

  7. .NET TCP协议之TcpClient与TcpListener交互

    问题:手机某项功能服务需要采用TCP协议与第三方交互通信.需先在公司内部测试此功能. 原因:第三方没有任何消息返回,也没有客服提供服务. 解决方法:公司内部做一个TCP协议服务器端,根据外网ip+端口 ...

  8. m序列c语言实现

    演示,不是算法 void m4() { int a[4]={1,0,0,1}; int m[15]; int temp; for(int i=0;i<15;i++){ m[i] = a[0]; ...

  9. 【Redis】Redis-benchmark测试Redis性能

    Redis-benchmark是官方自带的Redis性能测试工具,可以有效的测试Redis服务的性能. 使用说明如下: Usage: redis-benchmark [-h <host>] ...

  10. Django高级篇一RESTful架构及API设计

    一.什么是RESTful架构? 通过互联网通信,建立在分布式体系上"客户端/服务器模式”的互联网软件,具有高并发和高延时的特点. 简单的来说,就是用开发软件的模式开发网站.网站开发,完全可以 ...