codeforces 571B Minimization(dp)
2015-09-02 17:19
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OBLEMS
SUBMIT
CODE
MY
SUBMISSIONS
STATUS
HACKS
ROOM
STANDINGS
CUSTOM
INVOCATION
D. Minimization
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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.
Sample test(s)
input
output
input
output
input
output
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
计算|arr[1]-arr[k+1]|+|arr[2]-arr[k+2]|+......+|arr[n-k]-arr
|,不妨将前式转换为
|arr[1]-arr[k+1]|+|arr[k+1]-arr[k+1+k]|+|arr[k+1+k]-arr[k+1+k+k]|+.....(其中k+1+k+...+k<=数组元素的个数n)
+|arr[2]-arr[k+2]|+|arr[k+2]-arr[k+2+k]|+|arr[k+2+k]-arr[k+2+k+k]|+....(其中k+2+k+...+k<=数组元素的个数n)
+......
+|arr[k]-arr[k+k]|+|arr[k+k]-arr[k+k+k]|+|arr[k+k+k]-arr[k+k+k+k]|+....(其中k+k+k+...+k<=数组元素的个数n)
这样分成了以1,2....k开头的k组;每组的长度可能是n/k或者n/k+1;
长度为n/k+1的组数为n%k;长度为n/k的组数为k-n%k;
只有每组的元素都是有序的,每组的相邻元素的差的和才会最小;
若每组元素都是有序的,则上面k组的结果就分别等价于:
|arr[1]-arr[k+1+k+.....+k]| ,|arr[2]-arr[k+2+k+...k]| , ....... ,|arr[k]-arr[k+k+k+...k]|;
这样每组的和只与这一组的第一个元素和最后一个元素有关; 类似于((a1-a2)+(a2-a3)+...+(a[n-1]-a
))=a1-a
;
最终结果就是这k组的和;
所以现在的问题就是这k组中n%k个较长的组和k-n%k个较短的组怎么选;
比如第二组数据 -5,-5,3,3,3 ;其中长组的个数为1,短组的个数为1;段组的长度为2;长组的长度为3;
先对数据排序,如果选-5,-5,3为一组(长组),3,3为一组(短组);那么最后结果就是错的;
所以需要递推去分组;
我也是看了别人的程序后才彻底明白的,具体看代码;
SUBMIT
CODE
MY
SUBMISSIONS
STATUS
HACKS
ROOM
STANDINGS
CUSTOM
INVOCATION
D. Minimization
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard 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.
Sample test(s)
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
计算|arr[1]-arr[k+1]|+|arr[2]-arr[k+2]|+......+|arr[n-k]-arr
|,不妨将前式转换为
|arr[1]-arr[k+1]|+|arr[k+1]-arr[k+1+k]|+|arr[k+1+k]-arr[k+1+k+k]|+.....(其中k+1+k+...+k<=数组元素的个数n)
+|arr[2]-arr[k+2]|+|arr[k+2]-arr[k+2+k]|+|arr[k+2+k]-arr[k+2+k+k]|+....(其中k+2+k+...+k<=数组元素的个数n)
+......
+|arr[k]-arr[k+k]|+|arr[k+k]-arr[k+k+k]|+|arr[k+k+k]-arr[k+k+k+k]|+....(其中k+k+k+...+k<=数组元素的个数n)
这样分成了以1,2....k开头的k组;每组的长度可能是n/k或者n/k+1;
长度为n/k+1的组数为n%k;长度为n/k的组数为k-n%k;
只有每组的元素都是有序的,每组的相邻元素的差的和才会最小;
若每组元素都是有序的,则上面k组的结果就分别等价于:
|arr[1]-arr[k+1+k+.....+k]| ,|arr[2]-arr[k+2+k+...k]| , ....... ,|arr[k]-arr[k+k+k+...k]|;
这样每组的和只与这一组的第一个元素和最后一个元素有关; 类似于((a1-a2)+(a2-a3)+...+(a[n-1]-a
))=a1-a
;
最终结果就是这k组的和;
所以现在的问题就是这k组中n%k个较长的组和k-n%k个较短的组怎么选;
比如第二组数据 -5,-5,3,3,3 ;其中长组的个数为1,短组的个数为1;段组的长度为2;长组的长度为3;
先对数据排序,如果选-5,-5,3为一组(长组),3,3为一组(短组);那么最后结果就是错的;
所以需要递推去分组;
我也是看了别人的程序后才彻底明白的,具体看代码;
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<climits>
#include<list>
#include<stack>
#include<cmath>
#define ll long long
#define MAX 310000#define mem(a) memset(a,0,sizeof(a))
#define mems(a) memset(a,-1,sizeof(a))
using namespace std;
int arr[MAX];
ll dp[5500][5500];
int main()
{
int n,k;
scanf("%d%d",&n,&k);
int i;
for(i=1;i<=n;++i)
{
scanf("%d",&arr[i]);
}
sort(arr+1,arr+n+1);
int j;
dp[0][0]=0;
int num_long,num_short;
num_long=n%k;//较长组的个数
num_short=k-num_long;//较短组的个数
int len=n/k;//短组的长度
for(i=0;i<=num_long;++i)
{
for(j=0;j<=num_short;++j)
{
if(i==0&&j==0)
{
continue;
}
int pos=i*(len+1)+j*len;//当前分了多少个,也是当前所分的组的最后一个元素在arr中的坐标
dp[i][j]=INT_MAX;
if(j>0)
{
dp[i][j]=min(dp[i][j],dp[i][j-1]+arr[pos]-arr[pos-len+1]);
}
if(i>0)
{
dp[i][j]=min(dp[i][j],dp[i-1][j]+arr[pos]-arr[pos-len]);
}
}
}
printf("%lld\n",dp[num_long][num_short]);
}
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