题意：求大小为m的框内（从左向右滑动），最小值、最大值为多少
思路：维护一个单调队列，比如求最大值的时候，维护一个单调不递增队列，队首就是最大值（队首在框外的话，st++）。一个月前写过，竟然快要忘了
PS：G++交会TLE，用C++
#include<stdio.h> #include<iostream> #include<string.h> #include<string> #include<stdlib.h> #include<math.h> #include<vector> #include<list> #include<map> #include<stack> #include<queue> #include<algorithm> #include<numeric> #include<functional> using namespace std; typedef long long ll; const int maxn = 1e6+5; int que1[maxn],que2[maxn]; int a[maxn]; int ans1[maxn],ans2[maxn],tot1,tot2; int main(void) { int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int st1,en1,st2,en2; st1 = en1 = st2 = en2 = 0; tot1 = tot2 = 0; for(int i = 1; i <= n; i++) scanf("%d",&a[i]); for(int i = 1; i <= m; i++) { while(en2 > st2 && a[i] > a[ que2[en2-1] ]) en2--; que2[en2++] = i; while(en1 > st1 && a[i] < a[ que1[en1-1] ]) en1--; que1[en1++] = i; } ans2[tot2++] = a[que2[st2]]; ans1[tot1++] = a[que1[st1]]; for(int i = m+1; i <= n; i++) { if(que2[st2] < i-m+1) st2++; while(en2 > st2 && a[i] > a[ que2[en2-1] ]) en2--; que2[en2++] = i; if(que1[st1] < i-m+1) st1++; while(en1 > st1 && a[i] < a[ que1[en1-1] ]) en1--; que1[en1++] = i; ans2[tot2++] = a[que2[st2]]; ans1[tot1++] = a[que1[st1]]; } for(int i = 0; i < tot1; i++) printf("%d%c",ans1[i],i==tot1-1?'\n':' '); for(int i = 0; i < tot2; i++) printf("%d%c",ans2[i],i==tot2-1?'\n':' '); } return 0; }
