Whu oj Problem 1603 - Minimum Sum

最新推荐文章于 2016-07-18 17:54:26 发布

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最新推荐文章于 2016-07-18 17:54:26 发布

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本文介绍了一种解决特定数组问题的算法，目标是在指定数量的元素中找到和最小的组合。通过排序和动态计算，该算法有效地解决了这一挑战，适用于大规模数据集。

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Problem 1603 - Minimum Sum

Time Limit: 2000MS Memory Limit: 65536KB
Total Submit: 541 Accepted: 147 Special Judge: No

Description

There are n numbers A[1] , A[2] .... A[n], you can select m numbers of it A[B[1]] , A[B[2]] ... A[B[m]] ( 1 <= B[1] < B[2] .... B[m] <= n ) such that Sum as small as possible.

Sum is sum of abs( A[B[i]]-A[B[j]] ) when 1 <= i < j <= m.

Input

There are multiple test cases.
First line of each case contains two integers n and m.( 1 <= m <= n <= 100000 )
Next line contains n integers A[1] , A[2] .... A[n].( 0 <= A[i] <= 100000 )
It's guaranteed that the sum of n is not larger than 1000000.

Output

For each test case, output minimum Sum in a line.

Sample Input

4 2
5 1 7 10
5 3
1 8 6 3 10

Sample Output

2
8

开始想复杂了。先排序找到前m位的sum值，在一次向后移，其间记录最小值。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100010
using namespace std;
int a[maxn];
int n,m;
long long fun()//求前m位的sum值
{
    int tem=m-1;
    int cnt=1;
    long long sum=0;
    for(int i=1;i<m;++i)
    {
        int dif=a[i]-a[i-1];
        sum+=dif*(tem*cnt);
        tem--;
        cnt++;
    }
    return sum;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=0;i<n;++i)
            scanf("%d",&a[i]);
        sort(a,a+n);
        long long sum=fun();
        long long ans=sum;
        int k=0;
        for(int i=m;i<n;++i)
        {
            for(int j=k+1;j<i;++j)
            {
                sum-=(a[j]-a[k]);
                sum+=(a[i]-a[j]);
            }
            ans=min(ans,sum);
            k++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}