Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.

Output

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".

Sample Input

3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4

Sample Output

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible. 思路:完全背包问题,dp[i][j]表示前i种硬币,重量为j时达到的最小值,考虑最直观情况,dp[i][j]=min(dp[i-1][j-k*w]+k*c)(0<=k<=j/w),但此式在计算时有重复,例如在计算dp[i][j-w]时候,计算了dp[i-1][j-w-n*w]+n*c(1<=n<=(j/w)-1),也就是当1<=k时,以此类推,计算dp[i][j]时候就只需要算min(dp[i-1][j],dp[i][j-w])即可,相当于dp[i][j-w]已经把1<=k的答案算出来了,再用滚动数组优化即可
const int INF = 0x3f3f3f3f;

int dp[], w[], c[];

int main() {
ios::sync_with_stdio(false), cin.tie();
int T;
cin >> T;
while(T--) {
int E, F, V, N;
cin >> E >> F >> N;
V = F - E;
for(int i = ; i <= V; ++i) dp[i] = INF;
for(int i = ; i <= N; ++i) cin >> c[i] >> w[i];
for(int i = ; i <= N; ++i) {
for(int j = w[i]; j <= V; ++j)
dp[j] = min(dp[j], dp[j-w[i]]+c[i]);
}
if(dp[V] < INF) cout << "The minimum amount of money in the piggy-bank is " << dp[V] << ".\n";
else cout << "This is impossible.\n";
}
return ;
}
												

Day9 - D - Piggy-Bank POJ - 1384的更多相关文章

  1. poj 1384 Piggy-Bank(全然背包)

    http://poj.org/problem?id=1384 Piggy-Bank Time Limit: 1000MS Memory Limit: 10000K Total Submissions: ...

  2. POJ 1384 POJ 1384 Piggy-Bank(全然背包)

    链接:http://poj.org/problem?id=1384 Piggy-Bank Time Limit: 1000MS Memory Limit: 10000K Total Submissio ...

  3. POJ 1384 Piggy-Bank (ZOJ 2014 Piggy-Bank) 完全背包

    POJ :http://poj.org/problem?id=1384 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode ...

  4. POJ 1384 Intervals (区间差分约束,根据不等式建图,然后跑spfa)

    传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1384 Intervals Time Limit: 10000/5000 MS (Java/Others ...

  5. poj 1384 Piggy-Bank(完全背包)

    Piggy-Bank Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 10830   Accepted: 5275 Descr ...

  6. POJ 1384

    求猜存钱罐中至少有多少钱.容易知道金币总的重量,接着背包. #include<cstdio> #include<iostream> using namespace std; # ...

  7. POJ 1384 Piggy-Bank 背包DP

    所谓的全然背包,就是说物品没有限制数量的. 怎么起个这么intimidating(吓人)的名字? 事实上和一般01背包没多少差别,只是数量能够无穷大,那么就能够利用一个物品累加到总容量结尾就能够了. ...

  8. poj 1384完全背包

    题意:给出猪罐子的空质量和满质量,和n个硬币的价值和质量,求猪罐子刚好塞满的的最小价值. 思路:选择硬币,完全背包问题,塞满==初始化为无穷,求最小价值,min. 代码: #include<io ...

  9. ACM Piggy Bank

    Problem Description Before ACM can do anything, a budget must be prepared and the necessary financia ...

随机推荐

  1. Codeforces Round #611 (Div. 3) D

    There are nn Christmas trees on an infinite number line. The ii -th tree grows at the position xixi ...

  2. ConcurrentHashMap 实现缓存类

    参考:https://blog.csdn.net/woshilijiuyi/article/details/81335497 在规定时间内,使用 hashMap 实现一个缓存工具类,需要考虑一下几点 ...

  3. C++ STL之集合set的使⽤

    写在最前面,本文摘录于柳生笔记: set是集合,一个set里面个元素各不相同的,而且set会按照元素从小到大的进行排序,一下是set的常用方法:

  4. 【layui】提交表单

    1 <script type="text/javascript"> layui.use(['form', 'layer', 'jquery'], function () ...

  5. php对象、面向对象

    对象 万物皆对象 一切可见之物都是对象 一切不可见之物也是(抽象的事物也是对象): 对象包含两部分 1对象的组成元素 对象的数据模型又称为对象的属性,又被称为对象的成员变量 2.对象的行为 是对象的行 ...

  6. 反射实现定位Servlet中的方法

    public class BaseServlet extends HttpServlet{ @Override protected void service(HttpServletRequest re ...

  7. Bcrypt加密算法简介

    用户表的密码通常使用MD5等不可逆算法加密后存储,为防止彩虹表破解更会先使用一个特定的字符串(如域名)加密,然后再使用一个随机的salt(盐值)加密. 特定字符串是程序代码中固定的,salt是每个密码 ...

  8. redhat7.6 DNS配置正向解析

    1.安装DNS服务 yum install bind yum install bind-chroot 安装完的配置文件/etc/named.conf 启动systemctl start named.s ...

  9. 七 Struts2访问Servlet的API方式二:原生方式

    Struts2访问Servlet的API方式二:原生方式 和解耦合的方式不同,原生方式既可以拿到域对象,也可以调用域对象中的方法 前端jsp: <%@ page language="j ...

  10. 【剑指Offer面试编程题】题目1355:扑克牌顺子--九度OJ

    题目描述: LL今天心情特别好,因为他去买了一副扑克牌,发现里面居然有2个大王,2个小王(一副牌原本是54张^_^)...他随机从中抽出了5张牌,想测测自己的手气,看看能不能抽到顺子,如果抽到的话,他 ...