poj3016

题解

求n编的poj3666

然后dp

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=,M=;
int num[N],rt[N],size[N],T,tot,k,val[N],l[N],r[N],a[N],b[N],dist[N];
int c[N][],dp[N][],n,m,cost[M][M],cost1[M][M],cost2[M][M];
int merge(int x,int y)
{
    if (!x||!y)return x+y;
    if (val[x]<val[y])swap(x,y);
    c[x][]=merge(c[x][],y);
    size[x]=size[c[x][]]+size[c[x][]]+;
    if (dist[c[x][]]<dist[c[x][]])swap(c[x][],c[x][]);
    dist[x]=dist[c[x][]]+;
    return x;
}
void pop(int &x){x=merge(c[x][],c[x][]);}
int newnode(int x)
{
    size[++tot]=;
    val[tot]=x;
    c[tot][]=c[tot][]=dist[tot]=;
    return tot;
}
void makecost1()
{
    int res=,cnt=;
    for (int j=;j<=n;j++)
     {
         tot=res=cnt=;
        for (int i=j;i<=n;i++)
          {
             rt[++cnt]=newnode(b[i]);
             l[cnt]=r[cnt]=i;num[cnt]=;
             while (cnt>&&val[rt[cnt-]]>val[rt[cnt]])
               {
                 cnt--;
                 if (num[cnt+]&)res+=(val[rt[cnt]]-val[rt[cnt+]]);
                 num[cnt]+=num[cnt+];
                rt[cnt]=merge(rt[cnt],rt[cnt+]);
                r[cnt]=r[cnt+];
                while (size[rt[cnt]]*>r[cnt]-l[cnt]+)pop(rt[cnt]);
              }
             cost1[j][i]=res;
          }
     }
}
void makecost2()
{
    int res=,cnt=;
    for (int j=;j<=n;j++)
     {
         tot=res=cnt=;
        for (int i=j;i<=n;i++)
          {
             rt[++cnt]=newnode(b[i]);
             l[cnt]=r[cnt]=i;num[cnt]=;
             while (cnt>&&val[rt[cnt-]]>val[rt[cnt]])
               {
                 cnt--;
                 if (num[cnt+]&)res+=(val[rt[cnt]]-val[rt[cnt+]]);
                 num[cnt]+=num[cnt+];
                rt[cnt]=merge(rt[cnt],rt[cnt+]);
                r[cnt]=r[cnt+];
                while (size[rt[cnt]]*>r[cnt]-l[cnt]+)pop(rt[cnt]);
              }
             cost2[j][i]=res;
          }
     }
}
void doit()
{
    for (int i=;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i]-i;
    makecost1();
    for (int i=;i<=n;i++)b[i]=-a[i]-i;
    makecost2();
    memset(dp,0x3f,sizeof dp);
    dp[][]=;
    for (int i=;i<=n;i++)
     for (int j=i;j<=n;j++)cost[i][j]=min(cost1[i][j],cost2[i][j]);
    for (int i=;i<=n;i++)
     for (int j=;j<=k&&j<=i;j++)
      for (int p=j-;p<i;p++)
       dp[i][j]=min(dp[i][j],dp[p][j-]+cost[p+][i]);
    printf("%d\n",dp[n][k]);
}
int main()
{
    while (~scanf("%d%d",&n,&k),n|k)doit();
}