最最基础的单调队列题目。一个单增一个单减。还是可以借此好好理解一下单调队列的。
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 using namespace std; 5 6 #define maxx 1000005 7 8 int num[maxx], inque[maxx], dque[maxx], maxn[maxx], minn[maxx]; 9 int pre1, pre2, lst1, lst2; 10 int n, k; 11 12 int main() { 13 while(~scanf("%d%d", &n, &k)) { 14 pre1 = 0, pre2 = 0, lst1 = 0, lst2 = 0; 15 int cnt = 0; 16 for (int i=0; i<n; ++i) { 17 scanf("%d", &num[i]); 18 // 单增 19 while(pre1 < lst1 && num[inque[lst1-1]] > num[i]) 20 lst1--; 21 while(pre2 < lst2 && num[dque[lst2-1]] < num[i]) 22 lst2--; 23 inque[lst1++] = i; 24 dque[lst2++] = i; 25 while(pre1 < lst1 && i - inque[pre1] + 1 > k) 26 pre1++; 27 while(pre2 < lst2 && i - dque[pre2] + 1 > k) 28 pre2++; 29 if (i + 1 >= k) { 30 maxn[cnt] = num[inque[pre1]]; 31 minn[cnt] = num[dque[pre2]]; 32 cnt++; 33 } 34 } 35 for (int i=0; i<cnt-1; ++i) { 36 printf("%d ", maxn[i]); 37 } 38 printf("%d\n", maxn[cnt-1]); 39 for (int i=0; i<cnt-1; ++i) { 40 printf("%d ", minn[i]); 41 } 42 printf("%d\n", minn[cnt-1]); 43 } 44 return 0; 45 }
