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…

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(…

题意:给出价值为1,2,3,4,5,6的6种物品数量,问是否能将物品分成两份,使两份的总价值相等. 思路:求出总价值除二,做多重背包,需要二进制优化. 代码: #include<iostream> #include<cstdio> #include<cstring> using namespace std; int n[7]; int v,sum; bool flag; int dp[100000]; /*完全背包*/ void CompletePack(int cos…

多重背包模型  写的时候漏了一个等号找了半天 i<<=1 !!!!!! #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]; ]; ]; int main(){ ; ; ])==){ ]; ; ;j<=a[];j<<){ m[flag++]=j; a[]-=j; } ])m[flag++]=a[]; ;i<n;i++){ scanf(…

http://poj.org/problem?id=3260   Description Farmer John has gone to town to buy some farm supplies. Being a very efficient man, he always pays for his goods in such a way that the smallest number of coins changes hands, i.e., the number of coins he…

http://poj.org/problem?id=1787   描述 Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. Yo…

http://poj.org/problem?id=1742 n个硬币,面值分别是A1...An,对应的数量分别是C1....Cn.用这些硬币组合起来能得到多少种面值不超过m的方案. 多重背包,不过这题很容易超时,用背包九讲的代码有人说行,但是我提交还是超时,后来参考别人代码加了一些优化才能过,有时间要去搞清楚多重背包的单调队列优化. #include<cstdio> #include<cstring> #include<algorithm> using namespa…

POJ 2392 Space Elevator(贪心+多重背包) http://poj.org/problem?id=2392 题意: 题意:给定n种积木.每种积木都有一个高度h[i],一个数量num[i].另一个限制条件,这个积木所在的位置不能高于limit[i],问能叠起的最大高度? 分析: 本题是一道多重背包问题, 只是每一个物品的选择不只要受该种物品的数量num[i]限制, 且该物品还受到limit[i]的限制. 这里有一个贪心的结论: 我们每次背包选取物品时都应该优先放置当前limit…

转载地址:http://m.blog.csdn.net/blog/u010489766/9229011 题目链接:http://poj.org/problem?id=1276 题意:机器里面共有n种面额的钱币,每种各ni张,求机器吐出小于等于所要求钱币的最大值 解析1:这题大牛都用了多重背包,不过,我一同学想出了一种就这题而言特别简单有效的方 法.(话说我就认为这题本来就不需要用到背包的,因为n的范围只到10,太小了).方法就是对钱进行遍历,看这些钱一共能组成多少面额的钱,然后从 cash向下枚…