bzoj1177 [Apio2009]Oil

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1177

【题解】

发现分割方案就只有6种……

稍微分点类，然后大力算出以(i,j)为左上角/左下角/右上角/右下角的二维前缀/后缀内的max正方形即可。

然后大力分六种情况讨论一波，注意边界。

# include <stdio.h>
# include <string.h>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 1500 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, m, k, s[M][M];
int p[M][M];
int a[M][M], b[M][M], c[M][M], d[M][M], ans;

int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i=1; i<=n; ++i)
        for (int j=1, x; j<=m; ++j) {
            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)
            p[i][j] = s[i][j] - s[i-k][j] - s[i][j-k] + s[i-k][j-k];
    
    // 以(i,j)为右下角 
    for (int i=k; i<=n; ++i) 
        for (int j=k; j<=m; ++j) 
            a[i][j] = max(p[i][j], max(a[i-1][j], a[i][j-1]));
    
    // 以(i,j)为左下角 
    for (int i=k; i<=n; ++i)
        for (int j=m-k+1; j; --j) {
            int C = j+k-1; 
            b[i][j] = max(p[i][C], max(b[i-1][j], b[i][j+1]));
        }
    
    // 以(i,j)为右上角
    for (int i=n-k+1; i; --i) 
        for (int j=k; j<=m; ++j) {
            int C = i+k-1;
            c[i][j] = max(p[C][j], max(c[i+1][j], c[i][j-1]));
        }
    
    // 以(i,j)为左上角
    for (int i=n-k+1; i; --i)
        for (int j=m-k+1; j; --j) {
            int C1 = i+k-1, C2 = j+k-1;
            d[i][j] = max(p[C1][C2], max(d[i+1][j], d[i][j+1]));
        }
    
    // 两个竖线    
    for (int i=k; i<=n; ++i)
        for (int j=k+k; j<=m-k; ++j)
            ans = max(ans, p[i][j] + a[n][j-k] + b[n][j+1]);
    
    // 两个横线
    for (int i=k+k; i<=n-k; ++i)
        for (int j=k; j<=n; ++j) 
            ans = max(ans, p[i][j] + a[i-k][m] + c[i+1][m]);
    
    // 竖线，横线左
    for (int i=k; i<=n-k; ++i)
        for (int j=k; j<=m-k; ++j) 
             ans = max(ans, a[i][j] + c[i+1][j] + d[1][j+1]); 

    // 竖线，横线右
    for (int i=k; i<=n-k; ++i)
        for (int j=k; j<=m-k; ++j)
            ans = max(ans, a[n][j] + b[i][j+1] + d[i+1][j+1]);
    
    // 横线，竖线上
    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] + d[i+1][1]);
    
    // 横线，竖线下
    for (int i=k; i<=n-k; ++i)
        for (int j=k; j<=m-k; ++j) 
            ans = max(ans, a[i][m] + c[i+1][j] + d[i+1][j+1]); 
    
    printf("%d
", ans); 
    return 0;
}