Luogu 2216[HAOI2007]理想的正方形

Solution

二维单调队列， 这个数组套起来看得我眼瞎。。。

Code

#include<cstdio>
#include<algorithm>
#include<cstring>
#define rd read()
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define per(i,a,b) for(register int i = (a); i >= (b); --i)
#define Lb lb[i]
#define Rb rb[i]
#define Ls ls[i]
#define Rs rs[i]
using namespace std;

const int N = 1e3 + 100;
const int inf = ~0U >> 1;

int n, m, len, a[N][N];
int qb[N][N], lb[N], rb[N];
int qs[N][N], ls[N], rs[N];

int qB[N], LB, RB;
int qS[N], LS, RS;

int ans = inf;

int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

int main()
{
    n = rd; m = rd; len = rd;
    rep(i, 1, n) rep(j, 1, m) a[i][j] = rd;
    rep(i, 1, n) ls[i] = lb[i] = 1;
    rep(j, 1, m) {
        rep(i, 1, n) {
            while(qb[i][Lb] <= j - len && Lb <= Rb) Lb++;
            while(qs[i][Ls] <= j - len && Ls <= Rs) Ls++;

            while(Lb <= Rb && a[i][qb[i][Rb]] <= a[i][j]) Rb--;
            while(Ls <= Rs && a[i][qs[i][Rs]] >= a[i][j]) Rs--;
            qb[i][++Rb] = j;
            qs[i][++Rs] = j;
        }
        LS = LB = 1;
                RB = RS = 0;
                if(j >= len)
        rep(i, 1, n) {
            while(qB[LB] <= i - len && LB <= RB) LB++;
            while(qS[LS] <= i - len && LS <= RS) LS++;

            while(LB <= RB && a[qB[RB]][qb[qB[RB]][lb[qB[RB]]]] <= a[i][qb[i][Lb]]) RB--;
            while(LS <= RS && a[qS[RS]][qs[qS[RS]][ls[qS[RS]]]] >= a[i][qs[i][Ls]]) RS--;
            qB[++RB] = i;
            qS[++RS] = i;
                        int hb = qB[LB], hs = qS[LS];
                        if(i >= len) ans = min(ans, a[hb][qb[hb][lb[hb]]] - a[hs][qs[hs][ls[hs]]]);
        }
    }
    printf("%d
", ans);
}