bzoj 1177: [Apio2009]Oil

神啊！！

3个块有6种情况（网上神犇有图！），然后在6种情况里，分别找分出的3块的最大值就好。（发现可以用DP预处理一下分别以每个点给正方形的左下，左上，右下，右上为端点的最大值，然后处理答案的时候可以直接O(1)查询了）

（代码奇丑）

#include <bits/stdc++.h>
#define LL long long
#define lowbit(x) x&(-x)
#define inf 0x3f3f3f3f
#define N 2005
using namespace std;
inline int ra()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
int n,m,k,s[N][N],a[N][N],b[N][N],c[N][N],d[N][N],ans,t[N][N];
int main(int argc, char const *argv[])
{
    scanf("%d%d%d",&n,&m,&k);
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
        {
            int x; scanf("%d",&x);
            s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+x;
        }
    for (int i=k; i<=n; i++)
        for (int j=k; j<=m; j++)
            t[i][j]=s[i][j]-s[i-k][j]-s[i][j-k]+s[i-k][j-k];
    for (int i=1; i<=n; i++)
        for (int j=1; j<=n; j++)
            s[i][j]=t[i][j];
    for (int i=k; i<=n; i++)
        for (int j=k; j<=m; j++)
            a[i][j]=max(s[i][j],max(a[i-1][j],a[i][j-1]));
    for (int i=k; i<=n; i++)
        for (int j=m-k+1; j>=1; j--)
            b[i][j]=max(s[i][j+k-1],max(b[i][j+1],b[i-1][j]));
    for (int i=n-k+1; i>=1; i--)
        for (int j=k; j<=m; j++)
            c[i][j]=max(s[i+k-1][j],max(c[i+1][j],c[i][j-1]));
    for (int i=n-k+1; i>=1; i--)
        for (int j=m-k+1; j>=1; j--)
            d[i][j]=max(s[i+k-1][j+k-1],max(d[i+1][j],d[i][j+1]));
    for (int i=k; i<=n-k; i++)
        for (int j=k; j<=m-k; j++)
        {
            ans=max(ans,a[i][j]+b[i][j+1]+c[i+1][m]);
            ans=max(ans,a[i][m]+c[i+1][j]+d[i+1][j+1]);
            ans=max(ans,a[i][j]+d[1][j+1]+c[i+1][j]);
            ans=max(ans,a[n][j]+b[i][j+1]+d[i+1][j+1]);
        }
    for (int i=k+k; i<=n-k; i++)
        for (int j=k; j<=m; j++)
            ans=max(ans,s[i][j]+a[i-k][m]+d[i+1][1]);
    for (int i=k; i<=n; i++)
        for (int j=k+k; j<=m-k; j++)
            ans=max(ans,s[i][j]+a[n][j-k]+b[n][j+1]);
    cout<<ans;
    return 0;
}