HDU 2888 Check Corners(二维RMQ)

Check Corners

Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3063    Accepted Submission(s): 1100

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 44 4 10 72 13 9 115 7 8 2013 20 8 241 1 4 41 1 3 31 3 3 41 1 1 1

 

Sample Output

20 no13 no20 yes4 yes

 

Source

2009 Multi-University Training Contest 9 - Host by HIT 

 

题意：

给出一个矩阵，给你几个询问，每个询问问你子矩阵中的最大值，若最大值在角上，yes，否则no。

POINT：

二维的RMQ，之前没做过。

#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>using namespace std;const int N = 300+3;int n,m;int a[N][N];int dp[N][N][9][9];void prermq(){    for(int i=1;i<=n;i++)    {        for(int j=1;j<=m;j++)        {            dp[i][j][0][0]=a[i][j];        }    }    for(int a=0;1<<a<=n;a++)    {        for(int b=0;1<<b<=m;b++)        {            if(a==b&&b==0) continue;            for(int i=1;i-1+(1<<a)<=n;i++)            {                for(int j=1;j-1+(1<<b)<=m;j++)                {                    if(a==0)                    {                        dp[i][j][a][b]=max(dp[i][j][a][b-1],dp[i][j+(1<<(b-1))][a][b-1]);                    }                    else                        dp[i][j][a][b]=max(dp[i][j][a-1][b],dp[i+(1<<(a-1))][j][a-1][b]);                }            }        }    }}int query(int r1,int c1,int r2,int c2){    int k1=(int)log2((double)(r2-r1+1));    int k2=(int)log2((double)(c2-c1+1));    int a=dp[r1][c1][k1][k2];    int b=dp[r1][c2+1-(1<<k2)][k1][k2];    int c=dp[r2+1-(1<<k1)][c1][k1][k2];    int d=dp[r2+1-(1<<k1)][c2+1-(1<<k2)][k1][k2];    return max(max(a,b),max(c,d));}int ans;bool f(int x,int y){    if(a[x][y]==ans) return 1;    return 0;}int main(){    while(~scanf("%d %d",&n,&m))    {        memset(dp,0,sizeof dp);        memset(a,0,sizeof a);        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                scanf("%d",&a[i][j]);            }        }        prermq();        int q;        scanf("%d",&q);        while(q--)        {            int r1,c1,r2,c2;            scanf("%d %d %d %d",&r1,&c1,&r2,&c2);            ans=query(r1,c1,r2,c2);            printf("%d ",ans);            if(f(r1,c1)||f(r2,c1)||f(r1,c2)||f(r2,c2))            {                printf("yes\n");            }            else printf("no\n");        }            }}