RMQ

最新推荐文章于 2024-06-01 17:21:49 发布

wchyumo2009

最新推荐文章于 2024-06-01 17:21:49 发布

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分类专栏： RMQ 文章标签： RMQ

本文链接：https://blog.youkuaiyun.com/wchyumo2009/article/details/39995817

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本文介绍了一个解决HDU2888问题的算法实现，通过使用区间最大值查询(RMQ)来高效地处理矩阵中子矩形的最大值问题。该算法首先预处理所有可能范围的最大值，然后通过查询四个子区域的最大值来确定目标矩形的最大值。

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hdu 2888 check corners

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

#define MAXN 301

int dp[MAXN][MAXN][9][9];
int val[MAXN][MAXN];
int m, n;

void RMQ()
{
    int kn = int(trunc(log2((double)n))), km = int(trunc(log2((double)m)));
    for (int i = 0; i <= km; ++i){
        for (int j = 0; j <= kn; ++j){
            if (i == 0 && j == 0)continue;
            for (int r = 1; r + (1<<i) - 1 <= m; ++r){
                for (int c = 1; c + (1<<j) - 1 <= n; ++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 query(int r1, int c1, int r2, int c2)
{
    if (r1 > r2)swap(r1, r2);
    if (c1 > c2)swap(c1, c2);
    int km = int(trunc(log2((double)(r2-r1+1))));
    int kn = int(trunc(log2((double)(c2-c1+1))));
    int t1 = dp[r1][c1][km][kn];
    int t2 = dp[r2-(1<<km)+1][c1][km][kn];
    int t3 = dp[r1][c2-(1<<kn)+1][km][kn];
    int t4 = dp[r2-(1<<km)+1][c2-(1<<kn)+1][km][kn];
    return max(max(t1, t2), max(t3, t4));
}

int main()
{
    int q, r1, c1, r2, c2;
    while (~scanf ("%d%d", &m, &n)){
        memset (dp, 0, sizeof(dp));
        for (int i = 1; i <= m; ++i){
            for (int j = 1; j <= n; ++j){
                scanf ("%d", &dp[i][j][0][0]);
            }
        }
        RMQ();
        scanf ("%d", &q);
        while (q--){
            scanf ("%d%d%d%d", &r1, &c1, &r2, &c2);
            int ret = query(r1, c1, r2, c2);
            printf ("%d ", ret);
            if (ret == dp[r1][c1][0][0] || ret == dp[r1][c2][0][0] || ret == dp[r2][c1][0][0] || ret == dp[r2][c2][0][0]){
                printf ("yes\n");
            }
            else printf ("no\n");
        }
    }
    return 0;
}