zoukankan      html  css  js  c++  java
  • POJ 2019 Cornfields (二维RMQ)

    Cornfields
    Time Limit: 1000MS   Memory Limit: 30000K
    Total Submissions: 4911   Accepted: 2392

    Description

    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

    5 3 1
    5 1 2 6 3
    1 3 5 2 7
    7 2 4 6 1
    9 9 8 6 5
    0 6 9 3 9
    1 2
    

    Sample Output

    5
    

    Source

    二维RMQ。

    和一维的差不多。

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    int val[255][255];
    int mm[255];
    int dpmin[255][255][8][8];//最小值
    int dpmax[255][255][8][8];//最大值
    
    void initRMQ(int n,int m)
    {
        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= m;j++)
                dpmin[i][j][0][0] = dpmax[i][j][0][0] = val[i][j];
        for(int ii = 0; ii <= mm[n]; ii++)
            for(int jj = 0; jj <= mm[m]; jj++)
                if(ii+jj)
                    for(int i = 1; i + (1<<ii) - 1 <= n; i++)
                        for(int j = 1; j + (1<<jj) - 1 <= m; j++)
                        {
                            if(ii)
                            {
                                dpmin[i][j][ii][jj] = min(dpmin[i][j][ii-1][jj],dpmin[i+(1<<(ii-1))][j][ii-1][jj]);
                                dpmax[i][j][ii][jj] = max(dpmax[i][j][ii-1][jj],dpmax[i+(1<<(ii-1))][j][ii-1][jj]);
                            }
                            else
                            {
                                dpmin[i][j][ii][jj] = min(dpmin[i][j][ii][jj-1],dpmin[i][j+(1<<(jj-1))][ii][jj-1]);
                                dpmax[i][j][ii][jj] = max(dpmax[i][j][ii][jj-1],dpmax[i][j+(1<<(jj-1))][ii][jj-1]);
                            }
                        }
    }
    //查询矩形的最大值
    int rmq1(int x1,int y1,int x2,int y2)
    {
        int k1 = mm[x2-x1+1];
        int k2 = mm[y2-y1+1];
        x2 = x2 - (1<<k1) + 1;
        y2 = y2 - (1<<k2) + 1;
        return max(max(dpmax[x1][y1][k1][k2],dpmax[x1][y2][k1][k2]),max(dpmax[x2][y1][k1][k2],dpmax[x2][y2][k1][k2]));
    }
    //查询矩形的最小值
    int rmq2(int x1,int y1,int x2,int y2)
    {
        int k1 = mm[x2-x1+1];
        int k2 = mm[y2-y1+1];
        x2 = x2 - (1<<k1) + 1;
        y2 = y2 - (1<<k2) + 1;
        return min(min(dpmin[x1][y1][k1][k2],dpmin[x1][y2][k1][k2]),min(dpmin[x2][y1][k1][k2],dpmin[x2][y2][k1][k2]));
    }
    
    
    int main()
    {
        mm[0] = -1;
        for(int i = 1;i <= 500;i++)
            mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
        int N,B,K;
        while(scanf("%d%d%d",&N,&B,&K)==3)
        {
            for(int i = 1;i <= N;i++)
                for(int j = 1;j <= N;j++)
                    scanf("%d",&val[i][j]);
            initRMQ(N,N);
            int x,y;
            while(K--)
            {
                scanf("%d%d",&x,&y);
                printf("%d
    ",rmq1(x,y,x+B-1,y+B-1)-rmq2(x,y,x+B-1,y+B-1));
            }
        }
        return 0;
    }
  • 相关阅读:
    欢迎加入本人建的QQ群,讨论技术,生活及每天都有招聘信息
    ThreadPool(线程池) in .Net
    Android和Unity混合开发——解决方案
    HoloLens的显示分辨率有多少?
    ARGB_8888
    这几天用高通VUFORIA的体会
    利用WMI检测电脑硬件信息,没办法显示cpu的信息
    高版本teamview的成为被控制端时,会一直出现“正在初始化显示参数”
    Windows Server 2012 R2搭建IIS服务器
    如何设置Win7不待机 Win7进入待机状态会断网的解决方法
  • 原文地址:https://www.cnblogs.com/kuangbin/p/3227420.html
Copyright © 2011-2022 走看看