单调队列
把每列的最大值最小值预处理出来,压成一维用两个单调队列维护最大值最小值,和最左下表。
当队首元素相减不满足约束时出队,这个时候维护最小左下表,让它移动到两个队列中队首靠左的下表那继续更新答案。
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false)
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)){
w |= ch == '-', ch = getchar();
}
while(isdigit(ch)){
ret = (ret << 3) + (ret << 1) + (ch ^ 48);
ch = getchar();
}
return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
const int N = 1000;
int _, n, m, a[N][N], mx[N], mn[N], q1[N], q2[N], ans;
int main(){
for(_ = read(); _; _ --){
ans = 0;
n = read(), m = read();
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j ++) a[i][j] = read();
}
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= n; j ++){
mx[j] = 0, mn[j] = INF;
}
for(int t = i; t <= n; t ++){
for(int j = 1; j <= n; j ++){
mx[j] = max(mx[j], a[t][j]);
mn[j] = min(mn[j], a[t][j]);
}
int l1 = 1, r1 = 0, l2 = 1, r2 = 0, sgm = 1;
for(int j = 1; j <= n; j ++){
while(l1 <= r1 && mx[q1[r1]] <= mx[j]) r1 --;
q1[++r1] = j;
while(l2 <= r2 && mn[q2[r2]] >= mn[j]) r2 --;
q2[++r2] = j;
while(l1 <= r1 && l2 <= r2 && mx[q1[l1]] - mn[q2[l2]] > m){
if(q1[l1] == sgm) l1 ++;
if(q2[l2] == sgm) l2 ++;
sgm ++;
}
ans = max(ans, (j - sgm + 1) * (t - i + 1));
}
}
}
printf("%d
", ans);
}
return 0;
}