POJ 2823 双端队列

链接：

http://poj.org/problem?id=2823

题意：

给定一个长度为n的数列，a0,a1,···,an-1和一个整数k。

求数列bi=min{ai,ai+1,···,ai+k-1}(i=0,1,···,n-k)

和数列ci=max{ai,ai+1,···,ai+k-1}(i=0,1,···,n-k)

代码：

31 int a[MAXN], b[MAXN], c[MAXN];
32 deque<int> deq;
33
34 int main() {
35     int n, k;
36     cin >> n >> k;
37     rep(i, 0, n) scanf("%d", a + i);
38
39     rep(i, 0, n) {
40         while (!deq.empty() && a[*deq.rbegin()] >= a[i]) deq.pop_back();
41         deq.push_back(i);
42         if (i - k + 1 >= 0) {
43             b[i - k + 1] = a[*deq.begin()];
44             if (*deq.begin() == i - k + 1) deq.pop_front();
45         }
46     }
47     deq.clear();
48     rep(i, 0, n) {
49         while (!deq.empty() && a[*deq.rbegin()] <= a[i]) deq.pop_back();
50         deq.push_back(i);
51         if (i - k + 1 >= 0) {
52             c[i - k + 1] = a[*deq.begin()];
53             if (*deq.begin() == i - k + 1) deq.pop_front();
54         }
55     }
56
57     rep(i, 0, n - k + 1) printf("%d%c", b[i], i == n - k ? ‘\n‘ : ‘ ‘);
58     rep(i, 0, n - k + 1) printf("%d%c", c[i], i == n - k ? ‘\n‘ : ‘ ‘);
59     return 0;
60 }