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最新推荐文章于 2022-08-13 15:51:45 发布

原创最新推荐文章于 2022-08-13 15:51:45 发布 · 512 阅读

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RMQ（区间最大、小值）专栏收录该内容

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本文介绍了一种二维RMQ（区间最大值查询）问题的解决方案，通过预处理实现快速查询任意矩形区域内的最大值。核心在于利用动态规划的思想进行二维数组的最大值缓存，进而能在对数时间内完成查询。

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RMQ的二维应用，核心思想和一维一样

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN = 301;
const int MAXM = 301;
int n, m, val[MAXN][MAXM];
int dp[MAXN][MAXM][9][9];
int log ( int x )
{
    int ans = 0;
    while ( ( 1 << ans ) <= x ) { ans++; }
    return ans - 1;
}
void RMQ_2D_PRE()
{
    for ( int r = 1; r <= n; ++r )
    {
        for ( int c = 1; c <= m; ++c )
        {
            dp[r][c][0][0] = val[r][c];
        }
    }
    int mx = log ( n ), my = log ( m );
    for ( int i = 0; i <= mx; ++i )
    {
        for ( int j = 0; j <= my; ++j )
        {
            if ( i == 0 && j == 0 ) { continue; }
            for ( int r = 1; r + ( 1 << i ) - 1 <= n; ++r )
            {
                for ( int c = 1; c + ( 1 << j ) - 1 <= m; ++c )
                {
                    if ( i == 0 ) { dp[r][c][i][j] = max ( dp[r][c][i][j - 1], dp[r][c + ( 1 << ( j - 1 ) )][i][j - 1] ); }
                    else
                    {
                        dp[r][c][i][j] = max ( dp[r][c][i - 1][j], dp[r + ( 1 << ( i - 1 ) )][c][i - 1][j] );
                    }
                }
            }
        }
    }
}
int RMQ_2D ( int x1, int x2, int y1, int y2 )
{
	int kx = log(x2-x1+1);
	int ky = log(y2-y1+1);
	int m1 = dp[x1][y1][kx][ky];
	int m2 = dp[x2-(1<<kx)+1][y1][kx][ky];
	int m3 = dp[x1][y2-(1<<ky)+1][kx][ky];
	int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
	return max(max(m1, m2), max(m3, m4));
}
int main()
{
#ifdef  __GNUC__
    freopen ( "in.txt", "r", stdin );
#endif // __GNUC__
    int q;
    int x1, x2, y1, y2;
    while ( scanf ( "%d%d", &n, &m ) != EOF )
    {
        for ( int i = 1; i <= n; ++i )
        {
            for ( int j = 1; j <= m; ++j )
            {
                scanf ( "%d", &val[i][j] );
            }
        }
        RMQ_2D_PRE();
        scanf ( "%d", &q );
        while ( q-- )
        {
            scanf ( "%d%d%d%d", &x1, &y1, &x2, &y2 );
            int ans = RMQ_2D(x1, x2, y1, y2);
            printf("%d ", ans);
            if (ans == val[x1][y1] || ans == val[x1][y2] || ans == val[x2][y1] || ans == val[x2][y2])
				printf("yes\n");
			else
				printf("no\n");
        }
    }

    return 0;
}