M - 01背包 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u   Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become t…

题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87125#problem/M 题目: Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy…

A - Robberies Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2955 Appoint description: Description The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usu…

0 or 1 题目链接: Rhttp://acm.hust.edu.cn/vjudge/contest/122685#problem/R Description Given a n*n matrix C ij (1<=i,j<=n),We want to find a n*n matrix X ij (1<=i,j<=n),which is 0 or 1. Besides,X ij meets the following conditions: 1.X 12+X 13+...X 1…

  今天第一次接触了0/1背包问题,总结一下,方便以后修改.不对的地方还请大家不啬赐教! 上一个蓝桥杯的例题: 数据规模和约定 代码: import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); int m =sc.nextInt(); int[][] a =new…

0 or 1 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2811    Accepted Submission(s): 914 Problem Description Given a n*n matrix Cij (1<=i,j<=n),We want to find a n*n matrix Xij (1<=i,j<…

0,1背包问题:我写笔记风格就是想到哪里写哪里,有很多是旧的也没删除,代码内部可能有很多重复的东西,但是保证能运行出最后效果 '''学点高大上的遗传算法''' '''首先是Np问题的定义: npc:多项式复杂程度的非确定性问题, 首先是基本的0-1背包问题. ''' '''给定N个物品和一个背包,物品i的质量是Wi,其价值位Vi,背包的容量为C,问应该 如何选择装入背包的物品,使得转入背包的物品的总价值为最大? 在选择物品的时候,对每种物品i只有两种选择,即装入背包或不装入背包.不能将 物品i装…

摘要: 使用动态规划法求解0/1背包问题. 难度: 初级 0/1背包问题的动态规划法求解,前人之述备矣,这里所做的工作,不过是自己根据理解实现了一遍,主要目的还是锻炼思维和编程能力,同时,也是为了增进对动态规划法机制的理解和掌握. 值得提及的一个问题是,在用 JAVA 实现时, 是按算法模型建模,还是用对象模型建模呢? 如果用算法模型,那么 背包的值.重量就直接存入二个数组里:如果用对象模型,则要对背包以及背包问题进行对象建模.思来想去,还是采用了对象模型,尽管心里感觉算法模型似乎更好一些.有时…

Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10526    Accepted Submission(s): 3868 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能超过背包的容量1-P. 在这个条件下,让装入背包的物品的总价值,也就是bag[i].[v]的和最大 bag.v是每一件物品的价值,bag.p是每件物品的体积 像上面这样想是行不通的.下面有解释 这道题麻烦的是概率这东西没法用个循环表示出来,根据我以往的经验,指望着把给出的测试数据乘上一百或者一万这种…