hdu 2888(二维RMQ)

最新推荐文章于 2019-03-18 20:48:45 发布

原创最新推荐文章于 2019-03-18 20:48:45 发布 · 977 阅读

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本文介绍了一个关于二维RMQ（Range Maximum Query）问题的解决方法，通过构建预处理表格实现快速查询子矩阵最大值，并判断最大值是否位于子矩阵的四个角上。此方法适用于竞赛编程中涉及矩阵操作的问题。

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Check Corners

Time Limit: 2000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 858 Accepted Submission(s): 275

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

Sample Output

  
  
   
   20 no
13 no
20 yes
4 yes

Source

2009 Multi-University Training Contest 9 - Host by HIT

Recommend

gaojie

题目： http://acm.hdu.edu.cn/showproblem.php?pid=2888

分析：二维的RMQ问题，没有多大变化。。。

代码：

#include<cstdio>
#define max(a,b) (a>b?a:b)
#define mm 303
int f[mm][mm][9][9],s[mm][mm];
int n,m;
int log2(int x)
{
    int k=0;
    while((1<<(k+1))<x)++k;
    return k;
}
void ST()
{
    int i,j,u,v,logn=log2(n),logm=log2(m);
    for(u=0;u<=logn;++u)
        for(v=0;v<=logm;++v)
            if(u+v)for(i=1;i+(1<<u)-1<=n;++i)
                for(j=1;j+(1<<v)-1<=m;++j)
                if(v==0)f[i][j][u][v]=max(f[i][j][u-1][v],f[i+(1<<(u-1))][j][u-1][v]);
                else f[i][j][u][v]=max(f[i][j][u][v-1],f[i][j+(1<<(v-1))][u][v-1]);
}
int get(int r1,int c1,int r2,int c2)
{
    int k=log2(r2-r1+1),t=log2(c2-c1+1);
    int a=max(f[r1][c1][k][t],f[r1][c2-(1<<t)+1][k][t]);
    int b=max(f[r2-(1<<k)+1][c1][k][t],f[r2-(1<<k)+1][c2-(1<<t)+1][k][t]);
    return max(a,b);
}
void in(int &a)
{
    char c;
    while((c=getchar())<'0'||c>'9');
    for(a=0;c>='0'&&c<='9';c=getchar())a=a*10+c-'0';
}
void out(int x)
{
    if(x>9)out(x/10);
    putchar(x%10+48);
}
int main()
{
    int i,j,k,q,r1,r2,c1,c2;
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=1;i<=n;++i)
            for(j=1;j<=m;++j)in(s[i][j]),f[i][j][0][0]=s[i][j];
        ST();
        in(q);
        while(q--)
        {
            in(r1),in(c1),in(r2),in(c2);
            k=get(r1,c1,r2,c2);
            out(k);
            if(s[r1][c1]==k||s[r2][c2]==k||s[r1][c2]==k||s[r2][c1]==k)puts(" yes");
            else puts(" no");
        }
    }
    return 0;
}