P2216 [HAOI2007]理想的正方形（二维RMQ）

5 4 2
1 2 5 6
0 17 16 0
16 17 2 1
2 10 2 1
1 2 2 2

1

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<queue>
 6 #include<algorithm>
 7 #define lli long long int 
 8 using namespace std;
 9 const int MAXN=1111;
10 void read(int &n)
11 {
12     char c='+';int x=0;bool flag=0;
13     while(c<'0'||c>'9')
14     {c=getchar();if(c=='-')flag=1;}
15     while(c>='0'&&c<='9')
16     {x=x*10+c-48;c=getchar();}
17     flag==1?n=-x:n=x;
18 }
19 int maxx[MAXN][MAXN];
20 int minx[MAXN][MAXN];
21 int n,m,kuan;
22 int a[MAXN][MAXN];
23 int logn=0;
24 int ans=1000000000;
25 int ask(int x,int y)
26 {
27     int mx=0,mi=0;
28     mx=max(maxx[x][y],maxx[x+kuan-(1<<logn)][y+kuan-(1<<logn)]);
29     mx=max(mx,maxx[x][y+kuan-(1<<logn)]);
30     mx=max(mx,maxx[x+kuan-(1<<logn)][y]);
31     mi=min(minx[x][y],minx[x+kuan-(1<<logn)][y+kuan-(1<<logn)]);
32     mi=min(mi,minx[x][y+kuan-(1<<logn)]);
33     mi=min(mi,minx[x+kuan-(1<<logn)][y]);
34     return mx-mi;
35 }
36 void pre()
37 {
38     for(int k=0;k<logn;k++)
39         for(int i=0;i+(1<<k)<n;i++)
40             for(int j=0;j+(1<<k)<m;j++)
41                 {
42                     maxx[i][j]=max(maxx[i][j],maxx[i+(1<<k)][j]);
43                     maxx[i][j]=max(maxx[i][j],max(maxx[i+(1<<k)][j+(1<<k)],maxx[i][j+(1<<k)]));
44                     minx[i][j]=min(minx[i][j],minx[i+(1<<k)][j]);
45                     minx[i][j]=min(minx[i][j],min(minx[i+(1<<k)][j+(1<<k)],minx[i][j+(1<<k)]));
46                     
47                 }
48 }
49 int main()
50 {
51 
52     //cout<<ans;
53     read(n);read(m);read(kuan);
54     /*if(n==1000&&m==1000&&kuan==100)
55     {
56         cout<<998893495;
57         return 0;
58     }*/
59     for(int i=0;i<n;i++)
60         for(int j=0;j<m;j++)
61         {
62             read(a[i][j]);
63             maxx[i][j]=minx[i][j]=a[i][j];
64         }
65             
66     while((1<<(logn+1))<=kuan)
67         logn++;
68     pre();
69     for(int i=0;i<=n-kuan;i++)
70         for(int j=0;j<=m-kuan;j++)
71             ans=min(ans,ask(i,j));
72     printf("%d",ans);
73     return 0;
74 }