回来看看二中的域，顺手抢个最优解

41ms

Accepted
Status Time Cost Memory Cost
#1 Accepted 2ms 480.0 KiB
#2 Accepted 2ms 476.0 KiB
#3 Accepted 1ms 344.0 KiB
#4 Accepted 1ms 576.0 KiB
#5 Accepted 1ms 352.0 KiB
#6 Accepted 5ms 576.0 KiB
#7 Accepted 7ms 572.0 KiB
#8 Accepted 2ms 576.0 KiB
#9 Accepted 5ms 576.0 KiB
#10 Accepted 10ms 568.0 KiB

#include<bits/stdc++.h>

using namespace std;
int n,m,r,c,a[17][17],k[17],ans=INT_MAX;
int zz[17][17][17],ymq[17][17][17];
int d[17],sum[17][17],dp[17][17];
int in()
{
    int k=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9')
    {
        if(c=='-')f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9')k=k*10+c-'0',c=getchar();
    return k*f;
}
void out(int x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else out(x/10),putchar(x%10+'0');
}
void dfs(int dep)
{
    if(r+1-dep>n-k[dep-1]) return;
    if(dep==r+1)
    {
        for(int j=1;j<=m;j++)
        {
            d[j]=0;
            for(int i=1;i<=r;i++)
                d[j]+=zz[k[i]][k[i+1]][j];
        }
        for(int i=1;i<=m;i++)
            for(int j=i+1;j<=m;j++)
            {
                sum[i][j]=0;
                for(int t=1;t<=r;t++)
                    sum[i][j]+=ymq[i][j][k[t]];
            }
        for(int i=1;i<=m;i++)
            for(int j=1;j<=c;j++)
            {
                dp[i][j]=0x3f3f3f3f;
                for(int k=j-1;k<i;k++)
                    dp[i][j]=min(dp[i][j],dp[k][j-1]+sum[k][i]);
                dp[i][j]+=d[i];
                if(j==c&&c<=i&&i<=m)
                    ans=min(ans,dp[i][c]);
            }
        return;
    }
    for(int i=k[dep-1]+1;i<=n;i++)
    {
        k[dep]=i;
        dfs(dep+1);
    }
}
int main()
{
    n=in();m=in();r=in();c=in();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            a[i][j]=in();
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            for(int k=1;k<=m;k++)
                zz[i][j][k]=zz[j][i][k]=abs(a[i][k]-a[j][k]);
    for(int i=1;i<=m;i++)
        for(int j=i+1;j<=m;j++)
            for(int k=1;k<=n;k++)
                ymq[i][j][k]=ymq[j][i][k]=abs(a[k][i]-a[k][j]);
    dfs(1);
    out(ans);
    return 0;
}