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HDU_2888_Check Corners

2018-06-17 21:48:02来源：未知阅读 ()

Check Corners

Time Limit: 2000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3247 Accepted Submission(s): 1173

Problem Description

Paul draw a big m*n matrix A last month, whose entries Ai,j are all integer numbers ( 1 <= i <= m, 1 <= j <= n ). Now he selects some sub-matrices, hoping to find the maximum number. Then he finds that there may be more than one maximum number, he also wants to know the number of them. But soon he find that it is too complex, so he changes his mind, he just want to know whether there is a maximum at the four corners of the sub-matrix, he calls this “Check corners”. It’s a boring job when selecting too many sub-matrices, so he asks you for help. (For the “Check corners” part: If the sub-matrix has only one row or column just check the two endpoints. If the sub-matrix has only one entry just output “yes”.)

Input

There are multiple test cases.

For each test case, the first line contains two integers m, n (1 <= m, n <= 300), which is the size of the row and column of the matrix, respectively. The next m lines with n integers each gives the elements of the matrix which fit in non-negative 32-bit integer.

The next line contains a single integer Q (1 <= Q <= 1,000,000), the number of queries. The next Q lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner and lower-right corner of the sub-matrix in question.

Output

For each test case, print Q lines with two numbers on each line, the required maximum integer and the result of the “Check corners” using “yes” or “no”. Separate the two parts with a single space.

Sample Input

4 4 4 4 10 7 2 13 9 11 5 7 8 20 13 20 8 2 4 1 1 4 4 1 1 3 3 1 3 3 4 1 1 1 1

Sample Output

20 no 13 no 20 yes 4 yes

Source

2009 Multi-University Training Contest 9 - Host by HIT

二维区间max，打二维ST表
dp[i][j][e][f]表明从矩阵左上角(i,j)开始宽度范围是2^e，高度范围是2^f的矩形

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <cmath>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long           LL ;
15 typedef unsigned long long ULL ;
16 const int    maxn = 1e5 + 10   ;
17 const int    inf  = 0x3f3f3f3f ;
18 const int    npos = -1         ;
19 const int    mod  = 1e9 + 7    ;
20 const int    mxx  = 100 + 5    ;
21 const double eps  = 1e-6       ;
22 const double PI   = acos(-1.0) ;
23 
24 int max4(int a, int b, int c, int d){
25     return max(max(a,b),max(c,d));
26 }
27 int m, n, fac[9], dp[310][310][9][9];
28 int X1, Y1, X2, Y2, ans, mx, q;
29 int main(){
30     // freopen("in.txt","r",stdin);
31     // freopen("out.txt","w",stdout);
32     for(int i=0;i<9;i++)
33         fac[i]=(1<<i);
34     while(~scanf("%d %d",&m,&n)){
35         for(int i=1;i<=m;i++)
36             for(int j=1;j<=n;j++)
37                 scanf("%d",&dp[i][j][0][0]);
38         int rk=(int)(log((double)m)/log(2.0));
39         int ck=(int)(log((double)n)/log(2.0));
40         for(int e=0;e<=rk;e++)
41             for(int f=0;f<=ck;f++)
42                 if(e || f)
43                     for(int i=1;i+fac[e]-1<=m;i++)
44                         for(int j=1;j+fac[f]-1<=n;j++)
45                             if(!e)
46                                 dp[i][j][e][f]=max(dp[i][j][e][f-1],dp[i][j+fac[f-1]][e][f-1]);
47                             else if(!f)
48                                 dp[i][j][e][f]=max(dp[i][j][e-1][f],dp[i+fac[e-1]][j][e-1][f]);
49                             else
50                                 dp[i][j][e][f]=max4(dp[i][j][e-1][f-1],dp[i+fac[e-1]][j][e-1][f-1],dp[i][j+fac[f-1]][e-1][f-1],dp[i+fac[e-1]][j+fac[f-1]][e-1][f-1]);
51         scanf("%d",&q);
52         while(q--){
53             ans=0;
54             scanf("%d %d %d %d",&X1,&Y1,&X2,&Y2);
55             rk=(int)(log((double)(X2-X1+1))/log(2.0));
56             ck=(int)(log((double)(Y2-Y1+1))/log(2.0));
57             mx=max4(dp[X1][Y1][rk][ck],dp[X2-fac[rk]+1][Y1][rk][ck],dp[X1][Y2-fac[ck]+1][rk][ck],dp[X2-fac[rk]+1][Y2-fac[ck]+1][rk][ck]);
58             if(mx==dp[X1][Y1][0][0]||
59                 mx==dp[X1][Y2][0][0]||
60                 mx==dp[X2][Y1][0][0]||
61                 mx==dp[X2][Y2][0][0]){
62                 ans=1;
63             }
64             printf("%d %s\n",mx,ans?"yes":"no");
65         }
66     }
67     return 0;
68 }