多重背包模板- #include <stdio.h> #include <string.h> int a[7]; int f[100005]; int v, k; void ZeroOnePack(int cost, int weight) { for (int i = v; i >= cost; i--) if (f[i - cost] + weight > f[i]) f[i] = f[i - cost] + weight; } void CompletePack(…

Dividing Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 63980   Accepted: 16591 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbl…

Q: 倍增优化后, 还是有重复的元素, 怎么办 A: 假定重复的元素比较少, 不用考虑 Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the…

题意 有分别价值为1,2,3,4,5,6的6种物品,输入6个数字,表示相应价值的物品的数量,问一下能不能将物品分成两份,是两份的总价值相等,其中一个物品不能切开,只能分给其中的某一方,当输入六个0是(即没有物品了),这程序结束,总物品的总个数不超过20000 思路 裸的多重可行性背包,设dp[i]表示容量为i是否可装.状态设计看代码吧. 代码 [cpp] #include <iostream> #include <cstdio> #include <cmath> #in…

http://poj.org/problem?id=1014 题意:6个物品,每个物品都有其价值和数量,判断是否能价值平分. 思路: 多重背包.利用二进制来转化成0-1背包求解. #include<iostream> #include<string> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; ; int sum; ]; int d[max…

Dividing   Description Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they cou…

原题目:http://poj.org/problem?id=1014 题目大意: 有分别价值为1,2,3,4,5,6的6种物品,输入6个数字,表示相应价值的物品的数量,问一下能不能将物品分成两份,是两份的总价值相等,其中一个物品不能切开,只能分给其中的某一方,当输入六个0是(即没有物品了),这程序结束,总物品的总个数不超过20000 输出:每个测试用例占三行: 第一行: Collection #k: k为第几组测试用例 第二行:是否能分(具体形式见用例) 第三行:空白(必须注意,否则PE) 一:…

Dividing Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20635    Accepted Submission(s): 5813 Problem Description Marsha and Bill own a collection of marbles. They want to split the collection…

Description People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact pri…

题意: k个块,给出每个块的高度hi,数量ci,不能超过的高度: 求这些块可以组成的最大高度一个. 思路: 大致可看这个题是一个背包,背包的承重是高度. 对于每个物品,有他的价值是高度,还有限定的数量,看到这里就是一个多重背包, 然后对于每个物品还有一个限制是对于他的特定高度,这种怎么处理其实很简单吧. dp[]应该是一个存一个高度: wa了一发,没有考虑一维的时候更新要从小到大..所以按照特定高度升序一下就好了- -好菜啊 #include<cstdio> #include<iostr…