hdu-2888 Check Corners（二维RMQ模板题）

Check Corners

Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2424    Accepted Submission(s): 881

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
    
    
    
    
题意：n，m是矩阵的长宽，然后给出一个矩阵。 q次询问，每次给出矩阵的左上角坐标和右下角坐标，求在这个小矩阵里的最大值，还有这个最大值是否是矩阵四个角其中之一。
    
    
    
    
思路：普通的RMQ用来求一个区间的最大值，这里是平面问题，于是想到用二维RMQ来求四边形的最大值。写法和一维差不多，细节多注意下就好。
    
    
    
    
代码：
    
    
    
    
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; #define N 305 int ma[N][N]; int dp[N][N][9][9]; int maxs(int a,int b,int c,int d) {     int maxn=a;     if(b>maxn)         maxn=b;     if(c>maxn)         maxn=c;     if(d>maxn)         maxn=d;     return maxn; } void makermq(int m,int n) {     for(int i=0; i<m; i++)         for(int j=0; j<n; j++)             dp[i][j][0][0]=ma[i][j];     for(int q=0; (1<<q)<=m; q++)     {         for(int w=0; (1<<w)<=n; w++)         {             if(q==0&&w==0)                 continue;             for(int j=0; j+(1<<q)-1<m; j++)             {                 for(int i=0; i+(1<<w)-1<n; i++)                 {                     if(q==0)                     dp[j][i][q][w]=max(dp[j][i][q][w-1],dp[j][i+(1<<(w-1))][q][w-1]);                     else if(w==0)                     dp[j][i][q][w]=max(dp[j][i][q-1][w],dp[j+(1<<(q-1))][i][q-1][w]);                     else                     dp[j][i][q][w]=maxs(dp[j][i][q-1][w],dp[j+(1<<(q-1))][i][q-1][w],                                         dp[j][i][q][w-1],dp[j][i+(1<<(w-1))][q][w-1]);                 }             }         }     } } int finds(int x1,int y1,int x2,int y2) {     int k1=(int)(log(x2-x1+1.0)/log(2.0));     int k2=(int)(log(y2-y1+1.0)/log(2.0));     return maxs(dp[x1][y1][k1][k2],dp[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2],                dp[x1][y2-(1<<k2)+1][k1][k2],dp[x2-(1<<k1)+1][y1][k1][k2]); } int main() {     int n,m,q;     int x1,y1,x2,y2;     while(scanf("%d %d",&m,&n)!=EOF)     {         for(int i=0; i<m; i++)             for(int j=0; j<n; j++)                 scanf("%d",&ma[i][j]);         makermq(m,n);         scanf("%d",&q);         while(q--)         {             scanf("%d %d %d %d",&x1,&y1,&x2,&y2);             x1--;             y1--;             x2--;             y2--;             int ans=finds(x1,y1,x2,y2);             printf("%d ",ans);             if(ans==ma[x1][y1]||ans==ma[x1][y2]||ans==ma[x2][y1]||ans==ma[x2][y2])                 printf("yes\n");             else                 printf("no\n");         }     }     return 0; }