選擇了某個區間就必須選擇其所有子區間，容易想到這是一個最大權閉合子圖的模型。考慮將區間按長度分層，相鄰層按包含關係連邊，區間[i,j]的權值即d_i,j，其中最後一層表示長度為1的區間的同時也表示壽司本身，所以其權值減去x。這樣建出原圖，再用最大權閉合子圖的方法重建就行了。於是m=0的情況就解決了。給最後一層的點連向壽司代號，m=1也就做完了。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include
 
<algorithm>
using namespace std;
#define ll long long
#define N 110
#define M 6789
#define S 0
#define T 6666
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0
 
?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,a[N],v[N][N],id[N][N],id2[1010],p[M],t=-1,cnt,ans;
int d[M],cur[M],q[M];
 
struct data{int to,nxt,cap,flow;
}edge[M<<3];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,p[y]=t;
}
bool bfs()
{
    memset(d,255,sizeof(d));d[S]=0;
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
        {
            q[++tail]=edge[i].to;
            d[edge[i].to]=d[x]+1;
        }
    }while (head<tail);
    return ~d[T];
}
int work(int k,int f) 
{
    if (k==T) return f;
    int used=0;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+1==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
        edge[i].flow+=w,edge[i^1].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==0) d[k]=-1;
    return used;
}
void dinic()
{
    while (bfs())
    {
        memcpy(cur,p,sizeof(p));
        ans-=work(S,inf);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4873.in","r",stdin);
    freopen("bzoj4873.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=n;i++)
        for (int j=i;j<=n;j++)
        v[i][j]=read();
    memset(p,255,sizeof(p));
    for (int k=n;k>=1;k--)
        for (int i=1;i<=n-k+1;i++)
        {
            int j=i+k-1;
            id[i][j]=++cnt;
            if (i>1) addedge(id[i-1][j],id[i][j],inf);
            if (j<n) addedge(id[i][j+1],id[i][j],inf);
            int x=v[i][j];if (k==1) x-=a[i];
            if (x>0) ans+=x,addedge(S,id[i][j],x);
            else addedge(id[i][j],T,-x);
        }
    if (m==1)
    for (int i=1;i<=n;i++)
    {
        if (!id2[a[i]]) id2[a[i]]=++cnt,addedge(id2[a[i]],T,a[i]*a[i]);
        addedge(id[i][i],id2[a[i]],inf);
    }
    dinic();
    cout<<ans;
    return 0;
}