Codeforces 158B. Taxi 【贪心】

最新推荐文章于 2018-10-18 23:26:36 发布

原创最新推荐文章于 2018-10-18 23:26:36 发布 · 398 阅读

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本文介绍了一个基于贪心算法的解决方案，用于解决多个儿童小组前往聚会地点的问题。通过将小组按人数进行分组，并尽可能地让每辆出租车坐满，来最小化所需出租车的数量。

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http://codeforces.com/problemset/problem/158/B
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 si friends (1 ≤ si ≤ 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, …, sn (1 ≤ si ≤ 4). The integers are separated by a space, si 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.

Examples
inputCopy
5
1 2 4 3 3
output
4
inputCopy
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.
开始以为将每组人数排序后每四个放入即可。但是其实最贪心的做法就是2和2,3和1 4 尽量将每一辆车都坐满才是最优解

#include<iostream>
#include<algorithm>
using namespace std;
int group[100005];
int num[5];
int main()
{
    int n;
    cin>>n;
    int cnt = 0;
    for(int i = 0;i<n;i++)
    {
        cin>>group[i];
        num[group[i]]++;

    }
    cnt+=num[4];
    int temp1 = min(num[3],num[1]);
    num[3] = num[3] - temp1;
    num[1] = num[1] - temp1;
    cnt+=temp1;

    cnt +=num[3];//3已经不能和其他数字组合了只能一组乘坐一辆车

    cnt += num[2]/2; //人数为2的可以组成num[2]/2组

    num[2]-=num[2]/2*2;//剩下的人数为2的组的个数 1 or 0
    if(num[2]==1&&num[1]>=2)
    {
        cnt+=1;
        num[1]-=2;
    } 
    else if(num[2]==1&&num[1]<2)
    {
        cnt+=1;
        num[1] = 0;
    } 
    if (num[1]>0&&num[1]<=4) cnt+=1;//本来写的else if 但是多个else if 如果满足前面的一个后面的将不会判断 
    else if(num[1]>4)
    {
        if(num[1]%4==0)cnt = cnt+num[1]/4;
        else cnt=cnt+num[1]/4+1; 
    }
     cout<<cnt<<endl;
    return 0;
}