D - Cornfields POJ - 2019（二维ST表）

思路

算法讲解

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <queue>
#include <map>
/* #include <unordered_map> */
#include <bitset>
#include <vector>
void fre() { system("clear"), freopen("A.txt", "r", stdin); freopen("Ans.txt","w",stdout); }
void Fre() { system("clear"), freopen("A.txt", "r", stdin);}
#define ios ios::sync_with_stdio(false)
#define Pi acos(-1)
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define db double
#define Pir pair<int, int>
#define PIR pair<Pir, Pir>
#define m_p make_pair
#define INF 0x3f3f3f3f
#define esp 1e-7
#define mod (ll)(1e9 + 7)
#define for_(i, s, e) for(int i = (ll)(s); i <= (ll)(e); i ++)
#define rep_(i, e, s) for(int i = (ll)(e); i >= (ll)(s); i --)
#define sc scanf
#define pr printf
#define sd(a) scanf("%d", &a)
#define ss(a) scanf("%s", a)
using namespace std;
#define Max(a, b, c, d) max(max(a, b), max(c, d));
#define Min(a, b, c, d) min(min(a, b), min(c, d));

const int mxn = 255;
int mz[mxn][mxn];
int Log[mxn];
int st1[mxn][mxn][8][8];
int st2[mxn][mxn][8][8];
void init(int n)
{
    Log[1] = 0;
    for_(i, 2, n) Log[i] = Log[i / 2] + 1;
}

void ST(int n)
{
    for_(i, 1, n)
        for_(j, 1, n)
        st1[i][j][0][0] = st2[i][j][0][0] = mz[i][j];

    for(int k = 0; (1 << k) <= n; k ++)
        for(int l = 0; (1 << l) <= n; l ++)
        {
            if(l == 0 && k == 0) continue;
            for(int i = 1; i + (1 << k) - 1 <= n; i ++)
                for(int j = 1; j + (1 << l) - 1 <= n; j ++)
                {
                    if(k)
                    {
                        st1[i][j][k][l]=max(st1[i][j][k-1][l],st1[i+(1<<(k-1))][j][k-1][l]);
                        st2[i][j][k][l]=min(st2[i][j][k-1][l],st2[i+(1<<(k-1))][j][k-1][l]);
                    }
                    else
                    {
                        st1[i][j][k][l]=max(st1[i][j][k][l-1],st1[i][j+(1<<(l-1))][k][l-1]);
                        st2[i][j][k][l]=min(st2[i][j][k][l-1],st2[i][j+(1<<(l-1))][k][l-1]);
                    }
                }
        }
}


int main()
{
    /* fre(); */
    int n, m, q;
    sc("%d %d %d", &n, &m, &q);
    init(n);
    for_(i, 1, n)
        for_(j, 1, n)
        sd(mz[i][j]);
    ST(n);
    int l, r;
    int k = Log[m];
    while(q --)
    {
        sc("%d %d", &l, &r);
        int L = l + m - 1;
        int R = r + m - 1;
        int mx = Max(st1[l][r][k][k], st1[L - (1 << k) + 1][r][k][k], st1[l][R - (1 << k) + 1][k][k], st1[L - (1 << k) + 1][R - (1 << k) + 1][k][k]);
        int mn = Min(st2[l][r][k][k], st2[L - (1 << k) + 1][r][k][k], st2[l][R - (1 << k) + 1][k][k], st2[L - (1 << k) + 1][R - (1 << k) + 1][k][k]);
        pr("%d\n", mx - mn);
    }

    return 0;
}