2019牛客暑期多校训练营（第三场）- F Planting Trees

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\n", ans);
    }
    return 0;
}