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题意：和上一道一样，但是由于size变成了500，如果按照之前的做法会tle + mle…
很容易发现，由于是方阵，长宽是相等的，所以有一维是可以省略的。
也就是所谓的降维？
代码实现
1/* ***********************************************
2Author :111qqz
3Created Time :2016年05月16日 星期一 19时24分26秒
4File Name :code/loj/1081.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=501;
34int a[N][N];
35int dp[N][N][9];
36int n,q;
37
38void init_rmq()
39{
40 for ( int i = 1 ; i <= n ; i++)
41 for ( int j = 1 ; j <= n ; j++)
42 dp[i][j][0] = a[i][j];
43
44 for ( int i = 1 ; (1<<i)<= n ; i++)
45 for ( int p = 1 ; p + (1<<i)-1 <= n ; p++)
46 for ( int q = 1 ; q + (1<<i)-1 <= n ; q++)
47 dp[p][q][i] = max(max(dp[p][q][i-1],dp[p+(1<<(i-1))][q][i-1]),max(dp[p][q+(1<<(i-1))][i-1],dp[p+(1<<(i-1))][q+(1<<(i-1))][i-1]));
48
49
50
51}
52
53int rmq_max(int x1,int y1,int x2,int y2)
54{
55 int k = 0;
56 while (1<<(k+1)<= x2-x1+1) k++;
57
58 int tmp1 = dp[x1][y1][k];
59 int tmp2 = dp[x2-(1<<k)+1][y1][k];
60 int tmp3 = dp[x1][y2-(1<<k)+1][k];
61 int tmp4 = dp[x2-(1<<k)+1][y2-(1<<k)+1][k];
62
63 return max(max(tmp1,tmp2),max(tmp3,tmp4));
64}
65int main()
66{
67 #ifndef ONLINE_JUDGE
68 freopen("code/in.txt","r",stdin);
69 #endif
70
71 int T;
72 scanf("%d",&T);
73 int cas = 0 ;
74 while (T--)
75 {
76 printf("Case %d:\n",++cas);
77 scanf("%d %d",&n,&q);
78 ms(a,0);
79 for ( int i = 1 ; i <= n ; i++)
80 for ( int j = 1 ; j <= n ; j++) scanf("%d",&a[i][j]);
81
82 init_rmq();
83
84 while (q--)
85 {
86 int x,y,s;
87 scanf("%d %d %d",&x,&y,&s);
88
89 printf("%d\n",rmq_max(x,y,x+s-1,y+s-1));
90 }
91 }
92
93 #ifndef ONLINE_JUDGE
94 fclose(stdin);
95 #endif
96 return 0;
97}