C - Sequence
Crawling in process...Crawling failedTime Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Little Petya likes to play very much. And most of all he likes to play the following game:
He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.
The sequence a is called non-decreasing if a1 ≤ a2 ≤ ... ≤ aN holds, where N is the length of the sequence.
Input
The first line of the input contains single integer N (1 ≤ N ≤ 5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 109 by absolute value.
Output
Output one integer — minimum number of steps required to achieve the goal.
Sample Input
5 3 2 -1 2 11
4
5 2 1 1 1 1
1
///状态转移方程 f[i][j]=min(f[i-1][j]+abs(a[i]-b[j]),f[i][j-1])
///f[i][j]表示把前i个数变成递增序列，第i个数变成b[j]的最少步数。
///为什么把序列中的数变成原序列中的数能得到最优解，这个猜想还不太明白的说。
#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
long long f[5005];
int a[5005],b[5005];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<=n;i++)
f[i]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[i]=a[i];
}
sort(b+1,b+1+n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
f[j]+=abs(a[i]-b[j]);
if(j>1)
f[j]=min(f[j-1],f[j]);
}
cout<<f[n]<<endl;
}
}