Bookshelf 2
来源:互联网 发布:html源码大全 编辑:程序博客网 时间:2024/05/22 00:52
Bookshelf 2
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 91 Accepted Submission(s) : 41
Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.
FJ has N cows (1 ≤ N ≤ 20) each with some height of Hi (1 ≤Hi ≤ 1,000,000 - these are very tall cows). The bookshelf has a height ofB (1 ≤ B ≤ S, where S is the sum of the heights of all cows).
To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.
Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf. Your program should print the minimal 'excess' height between the optimal stack of cows and the bookshelf.
* Line 1: Two space-separated integers: N and B
* Lines 2..N+1: Line i+1 contains a single integer: Hi
* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.
5 1631356
1
package DynamicProgrammingPackage;import java.util.Scanner;/** * 0-1背包 */public class Bookshelf {public static int[] stateArr = null;public static int[] h = new int[20];//放牛的高度public static int n;//一共多少头牛public static int maxh = 0;public static int sum = 0;public static void main(String[] args) {Scanner sc = new Scanner(System.in);while(sc.hasNextInt()) {n = sc.nextInt();maxh = sc.nextInt();sum = 0;for(int i=0; i<n; i++) {h[i] = sc.nextInt();sum += h[i];}stateArr = new int[sum+1];dp();int result = min();min();System.out.println(result);}}private static int min() {for(int i=0; i<=sum; i++) {if(maxh <= stateArr[i]) {return stateArr[i] - maxh;}}return 0;}private static void dp() {for(int i=0; i<n; i++) {ZeroOnePack(h[i], h[i]);}}private static void ZeroOnePack(int cost, int weight) {for(int i=sum; i>=weight; i--) {stateArr[i] = Math.max(stateArr[i], stateArr[i-weight]+cost);}}}
- Bookshelf 2
- Bookshelf 2
- POJ3628 Bookshelf 2
- POJ 3628 - Bookshelf 2
- POJ 3628 - Bookshelf 2
- POJ 3628 Bookshelf 2
- poj-3628-Bookshelf 2
- POJ3628 Bookshelf 2
- poj3628-Bookshelf 2
- poj 3628 Bookshelf 2
- POJ 3628 Bookshelf 2
- POJ 3628 Bookshelf 2
- poj 3628 Bookshelf 2
- POJ 3628-Bookshelf 2
- POJ - 3628 Bookshelf 2
- POJ 3628 Bookshelf 2
- poj 3628 Bookshelf 2
- poj 3268 Bookshelf 2
- LabVIEW 程序中的线程 2 - LabVIEW 的执行系统 [编写高效率的代码]
- CString, char*, string的相互转换
- powerpoint2007中浏览网页
- x264新手入门完全指南
- LabVIEW 程序中的线程 3 - 线程的优先级 [编写高效率的代码]
- Bookshelf 2
- 高效代码审查的十个经验
- 异常信息获取观察
- 如何做好项目经理:大项目售前售后30种技巧(120页PPT讲义和思维导图免费分享)
- HDU1850-Being a Good Boy in Spring Festival
- LabVIEW 程序中的线程 4 - 动态连接库函数的线程 [编写高效率的代码]
- 单链表的逆转 C语言
- js将onclick 放到js文件中
- Windows7下修改驱动器号或卷标