hdu1158

hdu1158

DP.  用f[i][j]表示前i个月末人数为j时的最小开支.则f[i][j]=min{f[i-1][k]+cost(i,j)},其中k表示第i-1个月的人数,cost(i,j)为第i个月的开支.当k<=j时,cost(i,j)=hire*(j-k)+j*salary;当k>j时,cost(i,j)=fire*(k-j)+j*salary.详细见代码:

#include < iostream >
using   namespace  std;
const   int  inf = 100000000 ;
int  hire,fire,sal;
int  mon,men[ 13 ];
int  f[ 13 ][ 10000 ];
int  main()
{
    while(scanf("%d",&mon)!=EOF&&mon)
    {
        scanf("%d%d%d",&hire,&sal,&fire);
        int i,j,k,MAX_NUM=0;
        for(i=1;i<=mon;i++) 
        {
             scanf("%d",&men[i]);
             if(MAX_NUM<men[i]) MAX_NUM=men[i];
        }
        for(i=men[1];i<=MAX_NUM;i++)
            f[1][i]=hire*i+sal*i;
        for(i=2;i<=mon;i++)
            for(j=men[i];j<=MAX_NUM;j++)
            {
                int minn=inf;
                for(k=men[i-1];k<=MAX_NUM;k++)
                {
                    if(k<=j&&minn>f[i-1][k]+hire*(j-k)+j*sal)
                        minn=f[i-1][k]+hire*(j-k)+j*sal;
                    else if(k>j&&minn>f[i-1][k]+fire*(k-j)+j*sal)
                        minn=f[i-1][k]+fire*(k-j)+j*sal;
                }
                f[i][j]=minn;
            }
        int ans=inf;
        for(i=men[mon];i<=MAX_NUM;i++)
            if(f[mon][i]!=0&&f[mon][i]<ans)
                ans=f[mon][i];
        printf("%d\n",ans);
    }
    return 0;
}