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。

【code】

 1 # include <stdio.h>
 2 # include <string.h>
 3 # include <algorithm>
 4 // # include <bits/stdc++.h>
 5
 6 using namespace std;
 7
 8 typedef long long ll;
 9 typedef long double ld;
10 typedef unsigned long long ull;
11 const int M = 1500 + 10;
12 const int mod = 1e9+7;
13
14 # define RG register
15 # define ST static
16
17 int n, m, k, s[M][M];
18 int p[M][M];
19 int a[M][M], b[M][M], c[M][M], d[M][M], ans;
20
21 int main() {
22     scanf("%d%d%d", &n, &m, &k);
23     for (int i=1; i<=n; ++i)
24         for (int j=1, x; j<=m; ++j) {
25             scanf("%d", &x);
26             s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + x;
27         }
28     for (int i=k; i<=n; ++i)
29         for (int j=k; j<=m; ++j)
30             p[i][j] = s[i][j] - s[i-k][j] - s[i][j-k] + s[i-k][j-k];
31
32     // 以(i,j)为右下角
33     for (int i=k; i<=n; ++i)
34         for (int j=k; j<=m; ++j)
35             a[i][j] = max(p[i][j], max(a[i-1][j], a[i][j-1]));
36
37     // 以(i,j)为左下角
38     for (int i=k; i<=n; ++i)
39         for (int j=m-k+1; j; --j) {
40             int C = j+k-1;
41             b[i][j] = max(p[i][C], max(b[i-1][j], b[i][j+1]));
42         }
43
44     // 以(i,j)为右上角
45     for (int i=n-k+1; i; --i)
46         for (int j=k; j<=m; ++j) {
47             int C = i+k-1;
48             c[i][j] = max(p[C][j], max(c[i+1][j], c[i][j-1]));
49         }
50
51     // 以(i,j)为左上角
52     for (int i=n-k+1; i; --i)
53         for (int j=m-k+1; j; --j) {
54             int C1 = i+k-1, C2 = j+k-1;
55             d[i][j] = max(p[C1][C2], max(d[i+1][j], d[i][j+1]));
56         }
57
58     // 两个竖线
59     for (int i=k; i<=n; ++i)
60         for (int j=k+k; j<=m-k; ++j)
61             ans = max(ans, p[i][j] + a[n][j-k] + b[n][j+1]);
62
63     // 两个横线
64     for (int i=k+k; i<=n-k; ++i)
65         for (int j=k; j<=n; ++j)
66             ans = max(ans, p[i][j] + a[i-k][m] + c[i+1][m]);
67
68     // 竖线，横线左
69     for (int i=k; i<=n-k; ++i)
70         for (int j=k; j<=m-k; ++j)
71              ans = max(ans, a[i][j] + c[i+1][j] + d[1][j+1]);
72
73     // 竖线，横线右
74     for (int i=k; i<=n-k; ++i)
75         for (int j=k; j<=m-k; ++j)
76             ans = max(ans, a[n][j] + b[i][j+1] + d[i+1][j+1]);
77
78     // 横线，竖线上
79     for (int i=k; i<=n-k; ++i)
80         for (int j=k; j<=m-k; ++j)
81             ans = max(ans, a[i][j] + b[i][j+1] + d[i+1][1]);
82
83     // 横线，竖线下
84     for (int i=k; i<=n-k; ++i)
85         for (int j=k; j<=m-k; ++j)
86             ans = max(ans, a[i][m] + c[i+1][j] + d[i+1][j+1]);
87
88     printf("%d\n", ans);
89     return 0;
90 }