原題鏈接

題目描述：有一個ab的整數組成的矩陣，現請你從中找出一個nn的正方形區域，使得該區域所有數中的最大值和最小值的差最小。

輸入格式：第一行為3個整數，分別表示a,b,n的值第二行至第a+1行每行為b個非負整數，表示矩陣中相應位置上的數。每行相鄰兩數之間用一空格分隔。
100%的數據2<=a,b<=1000,n<=a,n<=b,n<=1000

輸出格式：僅一個整數，為ab矩陣中所有“nn正方形區域中的最大整數和最小整數的差值”的最小值。

輸入樣例：
5 4 2
1 2 5 6
0 17 16 0
16 17 2 1
2 10 2 1
1 2 2 2

輸出樣例：
1

解析：題意十分簡單，像是個二維的RMQ（可我不會打...）。

代碼如下：

#include<cstdio>
#include<algorithm>
using namespace std;

const int maxn = 1005;
int n, m, k, a[maxn][maxn], f[maxn][maxn], g[maxn][maxn], maxv[maxn][maxn], minv[maxn][maxn], ans;
int que[maxn], tail, head;

int read(void) {
    char c; while (c = getchar(), c < '0' || c >'9'); int x = c - '0';
    while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0'; return x;
}

int main() {
    n = read(); m = read(); k = read();
      for (int i = 1; i <= n; ++ i)
        for (int j = 1; j <= m; ++ j) a[i][j] = read();
      for (int i = 1; i <= n; ++ i) { //維護橫行的最大值 
        head = tail = 0; que[0] = 0;
        for (int j = 1; j <= m; ++ j) {
            if (a[i][j] <= a[i][que[tail - 1]]) que[tail ++] = j;
            else {
                while (a[i][j] > a[i][que[tail - 1]] && tail > head) tail --;
                que[tail ++] = j;
              }
            while (que[tail - 1] - que[head] + 1 > k) head ++;
            if (j >= k) f[i][j - k + 1] = a[i][que[head]];
          }
      }
      for (int j = 1; j <= m; ++ j) { //維護豎行的最大值 
        head = tail = 0; que[0] = 0;
        for (int i = 1; i <= n; ++ i) {
            if (f[i][j] <= f[que[tail - 1]][j]) que[tail ++] = i;
            else {
                while (f[i][j] > f[que[tail - 1]][j] && tail > head) tail --;
                que[tail ++] = i;
              }
            while (que[tail - 1] - que[head] + 1 > k) head ++;
            if (i >= k) maxv[i - k + 1][j] = f[que[head]][j];
          }
      }
      for (int i = 1; i <= n; ++ i) { //維護橫行的最小值 
        head = tail = 0; que[0] = 0;
        for (int j = 1; j <= m; ++ j) {
            if (a[i][j] >= a[i][que[tail - 1]]) que[tail ++] = j;
            else {
                while (a[i][j] < a[i][que[tail - 1]] && tail > head) tail --;
                que[tail ++] = j;
              }
            while (que[tail - 1] - que[head] + 1 > k) head ++;
            if (j >= k) g[i][j - k + 1] = a[i][que[head]];
          }
      }
      for (int j = 1; j <= m; ++ j) { //維護豎行的最小值 
        head = tail = 0; que[0] = 0;
        for (int i = 1; i <= n; ++ i) {
            if (g[i][j] >= g[que[tail - 1]][j]) que[tail ++] = i;
            else {
                while (g[i][j] < g[que[tail - 1]][j] && tail > head) tail --;
                que[tail ++] = i;
              }
            while (que[tail - 1] - que[head] + 1 > k) head ++;
            if (i >= k) minv[i - k + 1][j] = g[que[head]][j];
          }
      }
    ans = 2e9;
      for (int i = 1; i <= n - k + 1; ++ i)
        for (int j = 1; j <= m - k + 1; ++ j) ans = min(ans, maxv[i][j] - minv[i][j]);
    printf("%d", ans);
    return 0;
}
 

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