POJ 3628 Bookshelf 2 (01背包)

最新推荐文章于 2019-07-13 15:44:55 发布

转载最新推荐文章于 2019-07-13 15:44:55 发布 · 65 阅读

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原文链接：http://www.cnblogs.com/gcczhongduan/p/5094698.html

本文探讨如何通过01背包算法找到一组牛的高度之和，使得这组高度之和达到书架高度的最小可能值，并计算出最小值与书架高度之间的差值。详细解释了题目的输入和输出要求，提供了易于理解的代码实现。

Bookshelf 2

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7496		Accepted: 3451

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 H_i (1 ≤ H_i ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (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: H_i

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

Sample Output

Source

USACO 2007 December Bronze

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：这道题非常easy。是非常明显的01背包问题，这里的n个物品，每一个物品的重量为c[i]，价值为w[i]而且c[i]==w[i]，。容量为全部c[i]的和sum。仅仅要在f[]中从头開始找。找到一个最小>=b的就是题目要的解

一開始看错题了，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 10005
int 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;
}

转载于:https://www.cnblogs.com/gcczhongduan/p/5094698.html