1 //把一个序列转换成非严格递增序列的最小花费 POJ 3666 2 //dp[i][j]:把第i个数转成第j小的数，最小花费 3 4 #include <iostream> 5 #include <cstdio> 6 #include <cstdlib> 7 #include <algorithm> 8 #include <vector> 9 #include <math.h> 10 // #include <memory.h> 11 using namespace std; 12 #define LL long long 13 typedef pair<int,int> pii; 14 const int inf = 0x3f3f3f3f; 15 const LL MOD =100000000LL; 16 const int N = 3000+10; 17 const double eps = 1e-8; 18 void fre() {freopen("in.txt","r",stdin);} 19 void freout() {freopen("out.txt","w",stdout);} 20 inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;} 21 22 int a[N],b[N]; 23 LL dp[N][N]; 24 int main(){ 25 int n; 26 scanf("%d",&n); 27 for(int i=1;i<=n;i++){ 28 scanf("%d",&a[i]); 29 b[i]=a[i]; 30 } 31 sort(b+1,b+1+n); 32 for(int i=1;i<=n;i++){ 33 dp[1][i]=abs(a[1]-b[i]); 34 } 35 for(int i=2;i<=n;i++){ 36 LL minn=1e18; 37 for(int j=1;j<=n;j++){ 38 minn=min(minn,dp[i-1][j]); 39 dp[i][j]=minn+abs(a[i]-b[j]); 40 } 41 } 42 LL ans=1e18; 43 for(int i=1;i<=n;i++){ 44 ans=min(ans,dp[n][i]); 45 } 46 cout<<ans<<endl; 47 return 0; 48 }