poj_3627Bookshelf

最新推荐文章于 2019-07-13 15:50:24 发布

n3verl4nd

最新推荐文章于 2019-07-13 15:50:24 发布

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在有限的空间内，为了能够到达书架顶部，需要选择最少数量的奶牛进行堆叠以达到所需的高度。通过排序和贪心算法找到最优解。

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Bookshelf

Time Limit:1000MS		Memory Limit:65536K
Total Submissions:6316		Accepted:3155

Description

Farmer John recently bought a bookshelf for cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.

Each of theNcows (1 ≤N≤ 20,000) has some height ofH_i(1 ≤H_i≤ 10,000) and a total height summed across allNcows ofS. The bookshelf has a height ofB(1 ≤B≤S< 2,000,000,007).

To reach the top of the bookshelf taller than the tallest cow, 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. Since more cows than necessary in the stack can be dangerous, your job is to find the set of cows that produces a stack of the smallest number of cows possible such that the stack can reach the bookshelf.

Input

* Line 1: Two space-separated integers:NandB
* Lines 2..N+1: Linei+1 contains a single integer:H_i

Output

* Line 1: A single integer representing the size of the smallest set of cows that can reach the bookshelf.

Sample Input

Sample Output

排序后，贪心。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int cow[20005];
#pragma warning(disable : 4996)
int main()
{
	freopen("in.txt", "r", stdin);
	int n, b, i;
	cin >> n >> b;
	for(int i = 1; i <= n; i++)
	{
		cin >> cow[i];
	}
	sort(cow + 1, cow + n + 1);
	int sum = 0;
	for(i = n; i >= 1; i--)
	{
		sum += cow[i];
		if(sum >= b)
		{
			break;
		}
	}
	cout << n - i + 1 << endl;
}