P2216-[HAOI2007]理想的正方形

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 typedef double db;
 5 #define INF 0x3f3f3f3f
 6 #define _for(i,a,b) for(int i = (a);i < b;i ++)
 7 #define _rep(i,a,b) for(int i = (a);i > b;i --)
 8 inline ll read()
 9 {
10     ll ans = 0;
11     char ch = getchar(), last = ' ';
12     while(!isdigit(ch)) last = ch, ch = getchar();
13     while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
14     if(last == '-') ans = -ans;
15     return ans;
16 }
17 inline void write(ll x)
18 {
19     if(x < 0) x = -x, putchar('-');
20     if(x >= 10) write(x / 10);
21     putchar(x % 10 + '0');
22 }
23 int a,b,n;
24 int Log[1003];
25 int maxx[1003][1003][11];
26 int minn[1003][1003][11];
27 void searchLog()
28 {
29     Log[0] = -1;
30     _for(i,1,max(a,b)+1)
31         Log[i] = Log[i/2]+1;
32 }
33 int query(int x,int y)
34 {
35     int xl = x+n-1,yl = y+n-1;
36     int s = Log[n];
37     int MAX = max(maxx[x][y][s],max(maxx[xl-(1<<s)+1][yl-(1<<s)+1][s],
38     max(maxx[xl-(1<<s)+1][y][s],maxx[x][yl-(1<<s)+1][s])));
39     
40     int MIN = min(minn[x][y][s],min(minn[xl-(1<<s)+1][yl-(1<<s)+1][s],
41     min(minn[xl-(1<<s)+1][y][s],minn[x][yl-(1<<s)+1][s]))); 
42     return MAX-MIN;
43 }
44 int main()
45 {
46     a = read(),b = read(), n = read();
47     searchLog();
48     _for(i,1,a+1)
49         _for(j,1,b+1)
50             minn[i][j][0] = maxx[i][j][0] = read();
51     
52     _for(r,1,Log[n]+1)
53         for(int i = 1;i + (1<<r) - 1 <= a;i ++)
54             for(int j = 1;j + (1<<r) -1 <= b;j ++)
55             {
56                 maxx[i][j][r] = max(maxx[i][j][r-1],max(maxx[i][j+(1<<r-1)][r-1],
57                 max(maxx[i+(1<<r-1)][j][r-1],maxx[i+(1<<r-1)][j+(1<<r-1)][r-1])));
58                 
59                 minn[i][j][r] = min(minn[i][j][r-1],min(minn[i][j+(1<<r-1)][r-1],
60                 min(minn[i+(1<<r-1)][j][r-1],minn[i+(1<<r-1)][j+(1<<r-1)][r-1])));
61             }
62     
63     int ans = INT_MAX;
64     _for(i,1,a-n+2)
65         _for(j,1,b-n+2)
66             ans = min(ans,query(i,j));
67     write(ans);
68     return 0;
69 }
原文地址:https://www.cnblogs.com/Asurudo/p/11446892.html