【codeforces 571B】Minimization

最新推荐文章于 2025-02-25 19:32:43 发布

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最新推荐文章于 2025-02-25 19:32:43 发布

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原文链接：http://www.cnblogs.com/AWCXV/p/7626772.html

本文探讨了一道算法问题：给定一个整数数组及一个正整数k，通过对数组元素进行排列使得特定形式的和达到最小。文章详细介绍了如何通过动态规划方法解决此问题，并给出了解决方案的具体步骤。

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time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You’ve got array A, consisting of n integers and a positive integer k. Array A is indexed by integers from 1 to n.

You need to permute the array elements so that value

became minimal possible. In particular, it is allowed not to change order of elements at all.
Input
The first line contains two integers n, k (2 ≤ n ≤ 3·105, 1 ≤ k ≤ min(5000, n - 1)).

The second line contains n integers A[1], A[2], …, A[n] ( - 109 ≤ A[i] ≤ 109), separate by spaces — elements of the array A.

Output
Print the minimum possible value of the sum described in the statement.

Examples
input
3 2
1 2 4
output
1
input
5 2
3 -5 3 -5 3
output
0
input
6 3
4 3 4 3 2 5
output
3
Note
In the first test one of the optimal permutations is 1 4 2.

In the second test the initial order is optimal.

In the third test one of the optimal permutations is 2 3 4 4 3 5.

【题目链接】:http://codeforces.com/contest/572/problem/D

f[i][j]=min(f[i][j],f[i-1][j]+a[i*(t+1)+j*t]-a[(i-1)*(t+1)+j*t+1]);
f[i][j] = min(f[i][j],f[i][j-1]+a[i*(t+1)+j*t]-a[i*(t+1)+(j-1)*t+1]);

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 3e5+100;
const int MAXK = 5e3+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,k,a[MAXN],f[MAXK][MAXK];
int bn,sn,t;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rep1(i,0,MAXK-1)
        rep1(j,0,MAXK-1)
            f[i][j] = 2000000000;
    rei(n);rei(k);
    rep1(i,1,n)
        rei(a[i]);
    sort(a+1,a+1+n);
    t = n/k;
    bn = n%k,sn = k-bn;
    f[0][0] = 0;
    rep1(i,0,bn)
        rep1(j,0,sn)
            {
                if (i)
                    f[i][j] = min(f[i][j],f[i-1][j]+a[i*(t+1)+j*t]-a[(i-1)*(t+1)+j*t+1]);
                if (j)
                    f[i][j] = min(f[i][j],f[i][j-1]+a[i*(t+1)+j*t]-a[i*(t+1)+(j-1)*t+1]);
            }
    printf("%d\n",f[bn][sn]);
    return 0;
}