hdu 2888 Check Corners

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");elseprintf("no\n");        }    }    return 0;}