POJ - 2019(RMQ二维)

最新推荐文章于 2019-11-29 21:02:03 发布

夕林山寸

最新推荐文章于 2019-11-29 21:02:03 发布

阅读量438

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/bjfu170203101/article/details/87908181

本文深入探讨了二维RMQ（Range Maximum Query）算法的应用场景，特别是在处理大规模数据集中的最大值和最小值查询问题上。通过一个具体的问题背景——农场主FJ寻找最平坦土地种植玉米，介绍了如何使用二维RMQ算法高效解决查询问题。文章详细解释了算法的实现过程，包括数据预处理和查询操作，并提供了一段C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he's looking to build the cornfield on the flattest piece of land he can find.

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it.

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form "in this B x B submatrix, what is the maximum and minimum elevation?". The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield.

Input

* Line 1: Three space-separated integers: N, B, and K.

* Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc.

* Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1.

Output

* Lines 1..K: A single integer per line representing the difference between the max and the min in each query.

Sample Input

Sample Output

思路很简单，就直接二维RMQ，dp[k][j][i]第k行的rmq区间维护。

我竟然wa了8次。。。

死都没看到求mi时dpi【】【】【】少了个i。。。。。。。

好不容易碰到自己想出来的题。。wa在这种弱智的地方。。醉了。。。。

加油。。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define inf 0x3f3f3f3f
int dp[255][255][32];
int dpi[255][255][32];
int a[255][255];
void RMQ(int n)
{
    for(int k=1;k<=n;k++)
    for(int i=1;i<=n;i++)
        dpi[k][i][0]=dp[k][i][0]=a[k][i];
    for(int k=1;k<=n;k++)
    for(int j=1;(1<<j)<=n;j++)
        for(int i=1;i+(1<<j)-1<=n;i++)
        {
            dp[k][i][j]=max(dp[k][i][j-1],dp[k][i+(1<<j-1)][j-1]);
            dpi[k][i][j]=min(dpi[k][i][j-1],dpi[k][i+(1<<j-1)][j-1]);
        }
}
int main()
{
    int n,b,k,x,y;
    while(~scanf("%d%d%d",&n,&b,&k))
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        int len=0;
        while(1<<len+1<=b)
                len++;
        RMQ(n);
        while(k--)
        {
            scanf("%d%d",&x,&y);
            int mi,ma;
            mi=inf;
            ma=-inf;
        // cout<<ma<<"++"<<mi<<"--"<<endl;
            for(int i=x;i<x+b/*&&i<=n*/;i++)
            {
       //     if(f>n)f=n;
                ma=max(max(dp[i][y][len],dp[i][(y+b-1)-(1<<len)+1][len]),ma);
                mi=min(min(dpi[i][y][len],dpi[i][(y+b-1)-(1<<len)+1][len]),mi);
            }
      //  cout<<ma<<"++"<<mi<<endl;*/
            cout<<ma-mi<<endl;
        }
    }
    return 0;
}

人一我百，人百我万.夕林山寸，寻梦指尖！