传送门(<---可以点的) 描述 Bessie, like so many of her sisters, has put on a few too many pounds enjoying the delectable grass from Farmer John's pastures. FJ has put her on a strict diet of no more than H (5 <= H <= 45,000) kilograms of hay per day. Bessi…

链接:https://ac.nowcoder.com/acm/contest/1082/K 题目描述 Bessie, like so many of her sisters, has put on a few too many pounds enjoying the delectable grass from Farmer John's pastures. FJ has put her on a strict diet of no more than H (5 <= H <= 45,000)…

01背包... ----------------------------------------------------------------------- #include<cstdio> #include<algorithm> #include<cstring> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i < n ; i++ ) #define clr( x , c )…

3407: [Usaco2009 Oct]Bessie's Weight Problem 贝茜的体重问题 Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 48  Solved: 46[Submit][Status][Discuss] Description     贝茜像她的诸多姊妹一样,因为从约翰的草地吃了太多美味的草而长出了太多的赘肉．所以约翰将她置于一个及其严格的节食计划之中．她每天不能吃多过H(5≤日≤45000)公斤的干草．贝茜只能吃一整捆…

Description     贝茜像她的诸多姊妹一样,因为从约翰的草地吃了太多美味的草而长出了太多的赘肉．所以约翰将她置于一个及其严格的节食计划之中．她每天不能吃多过H(5≤日≤45000)公斤的干草．贝茜只能吃一整捆干草:当她开始吃一捆干草的之后就再也停不下来了．她有一个完整 的N(1≤N≤500)捆可以给她当作晚餐的干草的清单．她自然想要尽量吃到更多的干草．很自然地,每捆干草只能被吃一次(即使在列表中相同的重量可能出现2次,但是这表示的是两捆干草,其中每捆干草最多只能被吃掉一次)．    …

Knapsack problem Given a set of n items, each with a weight w[i] and a value v[i], determine a way to choose the items into a knapsack so that the total weight is less than or equal to a given limit B and the total value is as large as possible. Find…

题目链接:Knapsack problem 大意:给出T组测试数据,每组给出n个物品和最大容量w.然后依次给出n个物品的价值和体积. 问,最多能盛的物品价值和是多少? 思路:01背包变形,因为w太大,转而以v为下标,求出价值对应的最小体积,然后求出能够满足给出体积的最大价值. 经典题目,思路倒是挺简单的,就是初始化总觉得别扭...T_T大概,因为我要找的是最小值,所以初始化为maxn,就结了? 这个问题好像叫01背包的超大背包... 模拟一下样例吧! 1 5 15 // 初始化为dp[0] =…

Food Problem Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5445 Description Few days before a game of orienteering, Bell came to a mathematician to solve a big problem. Bell is preparing the dessert for the gam…

转化思维,把价值当成背包容量,选择最小的花费,从上到下枚举,找到当这个最小的花费. #include<iostream> #include<cstring> #include<cstdio> using namespace std; int main() { ],t,b,w[],v[],n; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&b); ; ;i <…

题目传送门 这题和01背包最大的区别在于它没有价值,所以我们可以人工给它赋一个价值,由于要求体积最大,把价值赋成体积即可.顺带一提,这题数据范围很大,二维会MLE,要压缩成一维才可以AC 下面给出参考程序: #include<iostream> using namespace std; ],v[],w[]; int main() { cin>>n>>m; ;i<=m;i++) { cin>>v[i]; w[i]=v[i]; } ;i<=m;i++…