P2216 [HAOI2007]理想的正方形(dp+单调队列优化)
题目链接:传送门
题目:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
题目描述 有一个a*b的整数组成的矩阵,现请你从中找出一个n*n的正方形区域,使得该区域所有数中的最大值和最小值的差最小。 输入输出格式 输入格式: 第一行为3个整数,分别表示a,b,n的值 第二行至第a+1行每行为b个非负整数,表示矩阵中相应位置上的数。每行相邻两数之间用一空格分隔。 输出格式: 仅一个整数,为a*b矩阵中所有“n*n正方形区域中的最大整数和最小整数的差值”的最小值。 输入输出样例 输入样例#1: 5 4 2 1 2 5 6 0 17 16 0 16 17 2 1 2 10 2 1 1 2 2 2 输出样例#1: 1 说明 问题规模 (1)矩阵中的所有数都不超过1,000,000,000 (2)20%的数据2<=a,b<=100,n<=a,n<=b,n<=10 (3)100%的数据2<=a,b<=1000,n<=a,n<=b,n<=100
思路:
用2*b个单调队列维护:
每列长度为n的最大值和最小值;
再用2个单调队列维护:
更新到当前行为止,行数为n的最大值的最大值,和最小值的最小值。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> using namespace std; const int MAX_N = 1e3 + 5; const int INF = 0x3f3f3f3f; struct Node{ int val, ind; Node(int v = 0, int i = 0) : val(v), ind(i) {} }querow[MAX_N], quecol[MAX_N][MAX_N], querow2[MAX_N], quecol2[MAX_N][MAX_N]; int a, b, n; int headrow, tailrow, headcol[MAX_N], tailcol[MAX_N]; int headrow2, tailrow2, headcol2[MAX_N], tailcol2[MAX_N]; int mat[MAX_N][MAX_N]; int main() { cin >> a >> b >> n; int ans = INF; for (int i = 1; i <= a; i++) for (int j = 1; j <= b; j++) scanf("%d", &mat[i][j]); for (int i = 1; i <= a; i++) headcol[i] = 1, tailcol[i] = 0, headcol2[i] = 1, tailcol2[i] = 0; for (int i = 1; i <= a; i++) { headrow = headrow2 = 1; tailrow = tailrow2 = 0; for (int j = 1; j <= b; j++) { while (headcol[j] <= tailcol[j] && quecol[j][tailcol[j]].val <= mat[i][j]) tailcol[j]--; quecol[j][++tailcol[j]] = Node(mat[i][j], i); while (headcol[j] <= tailcol[j] && quecol[j][headcol[j]].ind <= i-n) headcol[j]++; Node cur = quecol[j][headcol[j]]; while (headrow <= tailrow && querow[tailrow].val <= cur.val) tailrow--; querow[++tailrow] = Node(cur.val, j); while (headrow <= tailrow && querow[headrow].ind <= j-n) headrow++; Node _max = querow[headrow]; while (headcol2[j] <= tailcol2[j] && quecol2[j][tailcol2[j]].val >= mat[i][j]) tailcol2[j]--; quecol2[j][++tailcol2[j]] = Node(mat[i][j], i); while (headcol2[j] <= tailcol2[j] && quecol2[j][headcol2[j]].ind <= i-n) headcol2[j]++; Node cur2 = quecol2[j][headcol2[j]]; while (headrow2 <= tailrow2 && querow2[tailrow2].val >= cur2.val) tailrow2--; querow2[++tailrow2] = Node(cur2.val, j); while (headrow2 <= tailrow2 && querow2[headrow2].ind <= j-n) headrow2++; Node _min = querow2[headrow2]; if (i >= n && j >= n) ans = min(ans, _max.val - _min.val); } } cout << ans << endl; return 0; }