思路
- 二维ST表 维护二维子区间最值
代码
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <queue>
#include <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 = 301;
int Log[mxn];
int mz[mxn][mxn];
int st[mxn][mxn][9][9];
void init()
{
Log[1] = 0;
for_(i, 2, mxn - 1) Log[i] = Log[i / 2] + 1;
}
void ST(int n, int m)
{
for_(i, 1, n)
for_(j, 1, m) st[i][j][0][0] = mz[i][j];
for(int k = 0; (1 << k) <= n; k ++)
for(int l = 0; (1 << l) <= m; l ++)
{
if(! (k + l)) continue;
for(int i = 1; i + (1 << k) - 1 <= n; i ++)
for(int j = 1; j + (1 << l) - 1 <= m; j ++)
{
if(k) st[i][j][k][l] = max(st[i][j][k - 1][l], st[i + (1 << (k - 1))][j][k - 1][l]);
else st[i][j][k][l] = max(st[i][j][k][l - 1], st[i][j + (1 << (l - 1))][k][l - 1]);
}
}
}
int get_max(int r, int l, int R, int L)
{
int x = Log[R - r + 1];
int y = Log[L - l + 1];
return Max(st[r][l][x][y], st[R - (1 << x) + 1][l][x][y], st[r][L - (1 << y) + 1][x][y], st[R - (1 << x) + 1][L - (1 << y) + 1][x][y]);
}
bool check(int r, int l, int R, int L, int mx)
{
if(mz[r][l] == mx || mz[R][L] == mx || mz[R][l] == mx || mz[r][L] == mx)
return true;
return false;
}
int main()
{
init();
int n, m;
while(sc("%d %d", &n, &m) != EOF)
{
for_(i, 1, n)
for_(j, 1, m)
sd(mz[i][j]);
ST(n, m);
int q, l, r, L, R;
sd(q);
while(q --)
{
sc("%d %d %d %d", &r, &l, &R, &L);
int mx = get_max(r, l, R, L);
if(check(r, l, R, L, mx))
pr("%d yes\n", mx);
else
pr("%d no\n", mx);
}
}
return 0;
}