Bookshelf 2 ​ POJ - 3628

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title: Poj-3628
date: 2019-02-09 18:32:26
tags: 01背包

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Bookshelf 2

​ POJ - 3628

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 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: 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 16
3
1
3
5
6

Sample Output

1

农夫约翰有一队牛，每个牛有其自己的身高…，为了够到书架的最高层，可以让奶牛们叠罗汉，求奶牛们可以叠出的高度里比书架高而且余出高度最小的一个，输出差值。

01背包

将背包容量视为所有牛height之和，每头牛的height既是体积也是价值，计算完遍历一遍找到比书架高的第一个输出差值即可。

代码如下：

#include <cstdio>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
#define maxn 3500
#define INF 0x3f3f3f
#define push_back pb;
#define closein cin.tie(0)
#define debug cout<<"hello world!"<<endl;
typedef long long ll;
using namespace std;
int dp[10000005],a[25];
int main()
{
	closein;
	int n,m,sum;
	cin>>n>>m;
	sum=0;
	for(int i=0;i<n;i++) { cin>>a[i]; sum+=a[i]; }
	memset(dp,0,sizeof(dp));
	for(int i=0;i<n;i++)
	{
		for(int j=sum;j>=a[i];j--)
		{
			dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
		}
	}
	for(int i=0;i<=sum;i++)
	{
		if(dp[i]>=m)
		{
			cout<<dp[i]-m<<endl;
			break;
		}
	}
	return 0;
}

Status	Accepted
Time	204ms
Memory	39828kB
Length	695
Lang	G++

除此之外，这个问题很像排列组合，选取不同的奶牛组成不同高度，可以用二进制表示组合，然后挨个计算即可。

#include <cstdio>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
#define maxn 3500
#define INF 0x3f3f3f
#define push_back pb;
#define closein cin.tie(0)
#define debug cout<<"hello world!"<<endl;
typedef long long ll;
using namespace std;
int a[25],dp[1<<20];
int main()
{
	closein;
	int n,m;
	cin>>n>>m;
	for(int i=0;i<n;i++) cin>>a[i];
	memset(dp,0,sizeof(dp));
	for(int i=0;i< 1<<n;i++)
	{
		for(int j=0;j<20;j++)
		{
			int k=i>>j;
			if((k&1)==1) dp[i]+=a[j];
		}
	}
	sort(dp,dp+(1<<n));
	for(int i=0;i< 1<<n;i++)
	{
		if(dp[i]>=m)
		{
			cout<<dp[i]-m<<endl;
			break;
		}
	}
	return 0;
}

Status	Accepted
Time	344ms
Memory	4772kB
Length	693
Lang	G++

人生苦短，及时行乐！