HDU1114Piggy-Bank(完全背包)
2016-06-02 14:51
357 查看
Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19735 Accepted Submission(s): 10020
[align=left]Problem Description[/align]
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!
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.".
[align=left]Sample Input[/align]
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
[align=left]Sample Output[/align]
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.
/*题目大意:已知猪灌所能容纳的重量,然后告诉若干硬币的价值与重量。求使得用已知硬币装入猪灌
* 中使得猪灌中硬币价值总和最小 ,且要求猪灌必须被装满,若不能装满则输出 This is impossible.
*/
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int maxn = 999999;
#define mem(a) memset(a, 0, sizeof(a))
int dp[10010]; //dp[i]表示所装重量为i时候的最小价值
struct node {
int p, w;
}a[550];
int main() {
int t;
scanf("%d",&t);
while (t --) {
mem(a);
mem(dp);
int e, f;
scanf("%d%d",&e, &f);
e = f-e;
for (int i = 0; i<=e; i++) dp[i] = maxn;
dp[0] = 0;
int n;
scanf("%d",&n);
for (int i = 1; i<=n; i++) scanf("%d%d",&a[i].p, &a[i].w);
for (int i = 1; i<=n; i++) {
for (int j = a[i].w; j<=e; j++) {
dp[j] = min(dp[j], dp[j-a[i].w] + a[i].p);
}
}
if (dp[e] == maxn) printf("This is impossible.\n");
else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[e]);
}
return 0;
}
相关文章推荐
- lsof命令:查看Liunx进程所打开的文件命令
- require和include
- HDU1114Piggy-Bank(完全背包)
- 站立会议07(第二次冲刺)
- 安装新交互英语客户端提示找不到SOAP的解决方案
- iOS设置状态栏样式,statusBarStyle
- Vijos P1352 最大获利
- cms网站模板上传安装教程
- Struts2的国际化和访问资源文件的几种方式
- Linux 查找文件并删除文件内容
- leetcode No238. Product of Array Except Self
- mysql表空间被占用,同名表无法创建或导入
- MongoDB数据库启动和停止
- dubbo发布webservice服务
- Tomcat6.0的安装与配置(手把手教你)
- css布局
- linux常用命令
- Cocoapods第三方管理工具的安装
- windows下使用Critical Section和Mutex实现线程同步实例
- ThreadLocal的使用