icpc:problems:luogup1886
problems | |
---|---|
名称 | 滑动窗口 |
题目编号 | P1886 |
题目链接 | luogu.com.cn/… |
来源 | Luogu |
算法分类 | 队列, 单调队列 |
难易程度 | 容易 |
滑动窗口
想法
维护单调队列,当元素超过区间或元素的值没有竞争力时出队。
代码实现
#include <cstdio> #include <cstring> const int N = 1e6 + 10; int line[N], qu[N], sit[N], head = 0, last = -1; int main() { int n, k; scanf("%d %d", &n, &k); for (int i = 1; i <= n; i++) scanf("%d", &line[i]); for (int i = 1; i <= n; i++) { while (head <= last && qu[last] >= line[i]) last--; qu[++last] = line[i]; sit[last] = i; while (head <= last && sit[head] < i - k + 1) head++; if (i >= k) { printf("%d ", qu[head]); } } head = 0; last = -1; memset(qu, 0, sizeof(qu)); memset(sit, 0, sizeof(sit)); printf("\n"); for (int i = 1; i <= n; i++) { while (head <= last && qu[last] <= line[i]) last--; qu[++last] = line[i]; sit[last] = i; while (head <= last && sit[head] < i - k + 1) head++; if (i >= k) { printf("%d ", qu[head]); } } return 0; }
/app/www/public/data/pages/icpc/problems/luogup1886.txt · 最后更改: 2024/03/27 02:53 由 温婕莺