[BZOJ1047] [HAOI2007] Ideal square (monotone queue)

Description

There is a matrix of a*b integers, now ask you to find a n*n square area, so that the maximum and minimum values of all the numbers in the region
The difference is minimal.

Input

The first behavior is 3 integers, which represent the value of A,b,n the second row to the a+1 line each behavior b nonnegative integer, representing the number in the corresponding position in the matrix. Every
Rows are separated by a space between two adjacent numbers.
100% of Data 2<=a,b<=1000,n<=a,n<=b,n<=100

Output

Only one integer, which is the minimum value of the difference between the maximum and minimum integers in all the n*n square areas in the a*b matrix.

Sample Input5 4 2
1 2 5 6
0 17 16 0
16 17 2 1
2 10 2 1
1 2 2 2Sample Output1HINTSourceSolution

began to want to use two-dimensional $st$ table, check the puzzle found that there is less complexity,,,

Since the length and width of the matrix are limited to $n$, there is a sense of the maximum minimum value of the sliding window.

Calculates the maximum minimum value of each point starting at the left $n$ Point, and uses a monotone queue for the exercise to find the most value

Each point is then calculated as the maximum minimum value of the $n*n$ matrix in the lower-right corner, using the value just calculated, and using the monotone queue to find the most value for the column.

So we're out. The maximum value of the corresponding rectangle for each point minus the minimum value

1#include <bits/stdc++.h>2 using namespacestd;3 intq[1005], c[1005][1005], r[4][1005][1005];4 intMain ()5 {6     intA, B, N, front, back, ans =1000000000;7scanf"%d%d%d", &a, &b, &n);8      for(inti =1; I <= A; ++i)9          for(intj =1; J <= B; ++j)Tenscanf"%d", &c[i][j]); One      for(inti =1; I <= A; ++i) A     { -Front = back =0; -          for(intj =1; J <= B; ++j) the         { -             if(Front! = back && J-q[front +1] ==N) -++Front; -              while(Front! = back && C[i][j] <=C[i][q[back]]) +--Back ; -Q[++back] =J; +r[0][I][J] = C[i][q[front +1]]; A         } atFront = back =0; -          for(intj =1; J <= B; ++j) -         { -             if(Front! = back && J-q[front +1] ==N) -++Front; -              while(Front! = back && C[i][j] >=C[i][q[back]]) in--Back ; -Q[++back] =J; tor[1][I][J] = C[i][q[front +1]]; +         } -     } the      for(intj = N; J <= B; ++j) *     { $Front = back =0;Panax Notoginseng          for(inti =1; I <= A; ++i) -         { the             if(Front! = back && I-q[front +1] ==N) +++Front; A              while(Front! = back && r[0][I][J] <= r[0][q[back]][j]) the--Back ; +Q[++back] =i; -r[2][I][J] = r[0][q[front +1]][j]; $         } $Front = back =0; -          for(inti =1; I <= A; ++i) -         { the             if(Front! = back && I-q[front +1] ==N) -++Front;Wuyi              while(Front! = back && r[1][I][J] >= r[1][q[back]][j]) the--Back ; -Q[++back] =i; Wur[3][I][J] = r[1][q[front +1]][j]; -         } About     } $      for(inti = n; I <= A; ++i) -          for(intj = N; J <= B; ++j) -ans = min (ans, r[3][I][J]-r[2][i][j]); -printf"%d\n", ans); A     return 0; +}

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[BZOJ1047] [HAOI2007] ideal square (monotone queue)