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：

int a[MAXN], b[MAXN], c[MAXN];
deque<int> deq;

int main() {
    int n, k;
    cin >> n >> k;
    rep(i, 0, n) scanf("%d", a + i);

    rep(i, 0, n) {
        while (!deq.empty() && a[*deq.rbegin()] >= a[i]) deq.pop_back();
        deq.push_back(i);
        if (i - k + 1 >= 0) {
            b[i - k + 1] = a[*deq.begin()];
            if (*deq.begin() == i - k + 1) deq.pop_front();
        }
    }
    deq.clear();
    rep(i, 0, n) {
        while (!deq.empty() && a[*deq.rbegin()] <= a[i]) deq.pop_back();
        deq.push_back(i);
        if (i - k + 1 >= 0) {
            c[i - k + 1] = a[*deq.begin()];
            if (*deq.begin() == i - k + 1) deq.pop_front();
        }
    }

    rep(i, 0, n - k + 1) printf("%d%c", b[i], i == n - k ? '
' : ' ');
    rep(i, 0, n - k + 1) printf("%d%c", c[i], i == n - k ? '
' : ' ');
    return 0;
}