POJ 3628 Bookshelf 2 (01背包)
来源:互联网 发布:数据库设计经典案例 编辑:程序博客网 时间:2024/04/30 19:09
Description
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.
Input
* Line 1: Two space-separated integers: N and B
* Lines 2..N+1: Line i+1 contains a single integer: Hi
Output
* 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.
Sample Input
5 1631356
Sample Output
1
Source
题意:就是给出n和b,然后给出n个数,用这n个数中的某些,求出一个和,这个和是>=b的最小值,输出最小值与b的差。
分析:这道题很简单,是很明显的01背包问题,这里的n个物品,每个物品的重量为c[i],价值为w[i]并且c[i]==w[i],,容量为所有c[i]的和sum。只要在f[]中从头开始找,找到一个最小>=b的就是题目要的解
一开始看错题了,以为是最接近b的值,WA无数次。代码:46MS
#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>using namespace std;#define M 1000005#define N 10005int map[N],dp[M];int main(){ int i,j,n,v,sum; while(cin>>n>>v) { memset(dp,0,sizeof(dp)); memset(map,0,sizeof(map)); sum=0; for(i=1;i<=n;i++) {cin>>map[i];sum+=map[i];} for(i=1;i<=n;i++) for(j=sum;j>=map[i];j--) dp[j]=max(dp[j],dp[j-map[i]]+map[i]); for(i=1;i<=sum;i++) { if(dp[i]>=v) //明显,第一个比v大的一定满足条件。 {cout<<dp[i]-v<<endl;break;} } } return 0;}
- poj 3628 Bookshelf 2 01背包
- poj 3628 Bookshelf 2 01背包!!!
- poj 3628 Bookshelf 2 01背包
- POJ 3628 Bookshelf 2 (01背包)
- [01背包]POJ 3628 Bookshelf 2
- POJ 3628 Bookshelf 2(DP:01背包)
- poj 3628 Bookshelf 2 01背包
- POJ 3628 Bookshelf 2 (01背包)
- poj 3628 Bookshelf 2(01背包)
- poj 3628 Bookshelf 2 (01背包)
- 【POJ 3628】Bookshelf 2(01背包)
- POJ 3628 Bookshelf 2 (01背包)
- POJ-3628--Bookshelf 2--01背包
- poj 3628 Bookshelf 2(01背包)
- POJ 3628 Bookshelf 2(背包)
- poj 3628 Bookshelf 2 背包
- poj 3628 Bookshelf 2(01背包入门或者dfs)
- 简单01背包问题求解 POJ:3628 Bookshelf 2
- 冒泡排序(Java)
- Windows Phone 8.1 Update主要更新内容汇总
- hdu 2046 骨牌铺方格(递推)
- 第一篇博客
- hdu 1207 汉诺塔II
- POJ 3628 Bookshelf 2 (01背包)
- 使用STL的next_permutation函数生成全排列(C++)
- bzoj1050 [HAOI2006]旅行comf
- PL/SQL编程基础(章节摘要)
- hdu 2393 Higher Math
- (HTML学习)常用标签 超链接颜色控制
- HDUJ 1013 Digital Roots
- JavaFX文档(2)什么是JavaFX
- EBS R12多组织访问控制