Luogu 2216[HAOI2007]理想的正方形 - 单调队列

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 + ;
 const int inf = ~0U >> ;
 
 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 = , p = ; char c = getchar();
     for(; c > '' || c < ''; c = getchar()) if(c == '-') p = -;
     for(; c >= '' && c <= ''; c = getchar()) X = X *  + c - '';
     return X * p;
 }
 
 int main()
 {
     n = rd; m = rd; len = rd;
     rep(i, , n) rep(j, , m) a[i][j] = rd;
     rep(i, , n) ls[i] = lb[i] = ;
     rep(j, , m) {
         rep(i, , 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 = ;
                 RB = RS = ;
                 if(j >= len)
         rep(i, , 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\n", ans);
 }