poj 2019 Cornfields (二维rmq)

Posted by 111qqz on Monday, May 16, 2016

TOC

poj2019题目链接

题意:给一个方阵,k个查询,每个查询求某个方阵的最大值和最小值之差。

思路:二维rmq.同时用到最大值和最小值的话可以把初始化写在一起。

/* ***********************************************
Author :111qqz
Created Time :2016年05月16日 星期一 18时31分23秒
File Name :code/poj/2019.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=251;
int a[N][N];
int dp[N][N][8][8];
int dp2[N][N][8][8];
int n,b,q;

void init_rmq()
{
    for ( int i = 1 ;i  <= n ; i++)
    for ( int j = 1 ; j <= n ; j++)
        dp[i][j][0][0] = dp2[i][j][0][0] = a[i][j];

    
    for ( int i = 0 ; (1<<i)<= n ; i++)
    for ( int  j = 0 ;  (1<<j) <= n ; j++)
        if (i==0 && j==0) continue;
        else for ( int p = 1 ; p + (1<<i)-1 <= n ; p++)
            for ( int q = 1 ; q + (1<<j)-1 <= n ; q++)
            if (i==0)
            {
                dp[p][q][i][j] = max(dp[p][q][i][j-1],dp[p][q+(1<<(j-1))][i][j-1]);
                dp2[p][q][i][j] = min(dp2[p][q][i][j-1],dp2[p][q+(1<<(j-1))][i][j-1]);
            }
            else
            {
                dp[p][q][i][j] = max(dp[p][q][i-1][j],dp[p+(1<<(i-1))][q][i-1][j]);
                dp2[p][q][i][j] = min(dp2[p][q][i-1][j],dp2[p+(1<<(i-1))][q][i-1][j]);
            }
}


int _rmq(int x1,int y1,int x2,int y2)
{
    int k1 = 0 ;
    int k2 = 0 ;
    while (1<<(k1+1)<=x2-x1+1) k1++;
    while (1<<(k2+1)<=y2-y1+1) k2++;

    int tmp1 = dp[x1][y1][k1][k2];
    int tmp2 = dp[x2-(1<<k1)+1][y1][k1][k2];
    int tmp3 = dp[x1][y2-(1<<k2)+1][k1][k2];
    int tmp4 = dp[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2];
    
    int mx =  max(max(tmp1,tmp2),max(tmp3,tmp4));

    tmp1 = dp2[x1][y1][k1][k2];
    tmp2 = dp2[x2-(1<<k1)+1][y1][k1][k2];
    tmp3 = dp2[x1][y2-(1<<k2)+1][k1][k2];
    tmp4 = dp2[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2];

    int mn = min(min(tmp1,tmp2),min(tmp3,tmp4));

   // cout<<"mx:"<<mx<<" mn:"<<mn<<endl;

    return mx - mn;
}


int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    scanf("%d %d %d",&n,&b,&q);
    for ( int i = 1 ; i <= n ; i++)
        for ( int j = 1 ; j <= n ; j++) scanf("%d",&a[i][j]);
    init_rmq();

    while (q--)
    {
        int x1,y1;
        scanf("%d %d",&x1,&y1);
        printf("%d\n",_rmq(x1,y1,x1+b-1,y1+b-1));
    }



  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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