【优化&“调整”思想】VK Cup 2012 Qualification Round 1 / 158B Taxi(贪心)

最新推荐文章于 2021-10-22 21:45:13 发布

原创最新推荐文章于 2021-10-22 21:45:13 发布 · 1k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces #acm #贪心

acm之路--枚举/搜索/贪心/分治同时被 3 个专栏收录

100 篇文章

订阅专栏

Codeforces

83 篇文章

订阅专栏

算法详解&模板

70 篇文章

订阅专栏

本文介绍了一个来自 Codeforces 的编程挑战题目 B.Taxi 的解决方案。任务是计算最少需要多少辆出租车来运送分成不同小组的学校儿童到 Polycarpus 的生日聚会，每辆车最多可以容纳四名乘客，并且同一组的孩子必须乘坐同一辆车。

After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of s_i friends (1 ≤ s_i ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of groups of schoolchildren. The second line contains a sequence of integers s₁, s₂, ..., s_n (1 ≤ s_i ≤ 4). The integers are separated by a space, s_i is the number of children in the i-th group.

Output

Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.

   Sample test(s)
 
     input
   
5
1 2 4 3 3

     output
   
4

     input
   
8
2 3 4 4 2 1 3 1

     output
   
5

Note

In the first test we can sort the children into four cars like this:

the third group (consisting of four children),
the fourth group (consisting of three children),
the fifth group (consisting of three children),
the first and the second group (consisting of one and two children, correspondingly).

There are other ways to sort the groups into four cars.

water.

未优化：

/*62ms,400KB*/

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

int group[100001];///亦可以用multiset实现

int main()
{
	int n, count = 0, j = 0;
	scanf("%d", &n);
	for (int i = 0; i < n; ++i)
		scanf("%d", &group[i]);
	sort(group, group + n);
	for (int i = n - 1; i >= j; --i)
	{
		if (group[i] == 4)
			++count;
		else if (group[i] == 3)
		{
			if (group[j] == 1)
				++j;
            ++count;
		}
		else if (group[i] == 2)
		{
			if (group[j] == 1 && group[j + 1] == 1)
				j += 2;
			else if (i && group[i - 1] == 2)
				--i;
			else if (i - j == 1)
			{
			    ++count;
				break;
			}
            ++count;
		}
		else///group[i]==1
		{
			count += ceil((double)(i - j + 1) / 4);
			break;
		}
	}
	printf("%d", count);
	return 0;
}

优化：

/*62ms,0KB*/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cctype>

#include <cmath>
#include <algorithm>

#include <vector>
#include <list>
#include <deque>
#include <stack>
#include <queue>
#include <map>
#include <set>

typedef unsigned long long ULL;
typedef long long LL;
using namespace std;

int arr[5];

int main()
{
	int N, id;
	scanf("%d", &N);
	for (int i = 0; i < N; ++i)
	{
		scanf("%d", &id);
		arr[id]++;
	}
	int ANS = arr[4] + arr[3];///先把最初能加的都加上
	///“调整”思想
	if (arr[1] > arr[3])
		arr[1] -= arr[3];
	else
		arr[1] = 0;
	if (arr[2] % 2 == 0)
		ANS += (arr[2] / 2);
	else
	{
		ANS += (arr[2] / 2) + 1;
		arr[1] -= 2;
	}
	if (arr[1] > 0)
		ANS += arr[1] / 4 + (arr[1] % 4 ? 1 : 0);
	printf("%d", ANS);
	return 0;
}