BZOJ 1592 DP

思路：f[i][j]表示前i个数 最后一个数是原数列的第j个的 最小值

//By SiriusRen#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n,a[2222],cpy[2222],minn[2222],f[2222][2222],temp[2222],ans=0x7fffffff;void work(){    memset(f,0x3f,sizeof(f));    memset(minn,0,sizeof(minn));    for(int i=1;i<=n;i++){        memset(temp,0x3f,sizeof(temp));        for(int j=1;j<=n;j++){            f[i][j]=min(minn[j]+abs(cpy[j]-a[i]),f[i][j]);            if(j==1)temp[j]=f[i][j];            else temp[j]=min(temp[j-1],f[i][j]);        }        for(int j=1;j<=n;j++)minn[j]=temp[j];    }    for(int i=1;i<=n;i++)ans=min(ans,f[n][i]);}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)scanf("%d",&a[i]),cpy[i]=a[i];    sort(cpy+1,cpy+1+n);    work(),reverse(a+1,a+1+n),work();    printf("%d\n",ans);}