SDNU 1501.Problem_J 贪心

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解决一个关于如何将一块整肉切割成N份指定大小小块肉，并使总消耗能量最小的问题。采用贪心策略，通过排序和逐步合并最小单位的方法找到最优解。

1501.Problem_J
Time Limit: 2000 MS Memory Limit: 32768 KB

Description

So long coding time make ChaoChao be more nonsense. In order to spend the boring time, ChaoChao plan to practice his cooking stills.

Now, CC wants to make N foods. The i-th food need a[i] unit meat, but CC find that there is only a whole meat and he need to consume energy to cut meat. Consumption value is equal to the size of the cut meat. For example, there is a 21 units meat, so CC will consume 21 energy to cut it. CC wants to cut meat into N small meat. He wants to know the minimum number of the energy he need.(The size of the meat bought is equal to the sum of N pieces of meat.)

INPUT

The first line is a number N (1 ≤ N ≤ 20000) identified the number of small meat.

Then next N lines, each line contains one number a[i](1 ≤ a[i] ≤ 50000) means a[i] units of the i-th meat.

Input

The first line is a number N (1 ≤ N ≤ 20000) identified the number of small meat.

Then next N lines, each line contains one number a[i](1 ≤ a[i] ≤ 50000) means a[i] units of the i-th meat.

Output

The minimum number of the energy.

Sample Input

3
8
5
8

Sample Output

34

一开始看完这个题就当做是石子合并类的题了，先排个序然后直接往上加，结果就是收获了一堆wa→_→后来才发现这种做法不行，这题也是比完赛之后才明白的怎么做。直接贪心处理，把这些数进行排序，然后从最小的两个合并，再进行排序，再合并最小的两个，以此类推直到所有的合并在了一起。就相当于把切肉的这个过程给倒过来运行了。

要注意的有两点，一开始我是每一遍循环都sort了一下，然后就TLE了，所以优化了一下，只sort一次，然后里面排序直接排合并的那个数就可以了。另外就是要用long long来处理这道题。

下面是AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long a[20005];

int cmp(long long a,long long b)
{
    return a>b;
}

int main()
{
    int n;
    int i,j;
    long long ans;
    long long t;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
        }
        sort(a,a+n,cmp);
        for(i=0;i<n-1;i++)
        {
            for(j=n-i-1;j>0;j--)
            {
                if(a[j]>a[j-1])
                {
                    t=a[j];
                    a[j]=a[j-1];
                    a[j-1]=t;
                }
                else
                    break;
            }
            a[n-i-2]=a[n-i-1]+a[n-i-2];
            ans+=a[n-i-2];
        }
        cout<<ans<<endl;
    }
    return 0;
}