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.
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
- 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 %sn",mx,ans?"yes":"no"); 65 } 66 } 67 return 0; 68 }
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