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jzoj 1326

题目大意

给你一个序列a和一个数k，让你求a中所有长为k的子序列的最大值和最小值

输入样例

8 3
1 3 -1 -3 5 3 6 7

输出样例

-1 -3 -3 -3 3 3
3 3 5 5 6 7

数据范围
$20$
$100$

解题思路

方法一：
滚动ST表（不多做解释）

代码1:

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long 
using namespace std;
int n, k, m, r, f[2][1000010], mf[2][1000010];
int main()
{memset(mf, 127/3, sizeof(mf));memset(f, -127/3, sizeof(f));scanf("%d %d", &n, &k);for (int i = 1; i <= n; ++i){scanf("%d", &f[0][i]);mf[0][i] = f[0][i];}m = log2(k);for (int j = 1; j <= m; ++j){memset(mf[j&1], 127/3, sizeof(mf[j&1]));//滚动memset(f[j&1], -127/3, sizeof(f[j&1]));for (int i = 1; i <= n - (1<<j) + 1; ++i){f[j&1][i] = max(f[(j + 1)&1][i], f[(j + 1)&1][i + (1<<(j - 1))]);//st表mf[j&1][i] = min(mf[(j + 1)&1][i], mf[(j + 1)&1][i + (1<<(j - 1))]);} }for (int i = 1; i <= n - k + 1; ++i){r = i + k - 1;printf("%d ", min(mf[m&1][i], mf[m&1][r - (1<<m) + 1]));//求解}putchar(10);for (int i = 1; i <= n - k + 1; ++i){r = i + k - 1;printf("%d ", max(f[m&1][i], f[m&1][r - (1<<m) + 1]));}return 0;
} 

---

方法二：
我们用单调队列来存最大/小值，如果不是更优的就丢掉，如果超过了也丢掉

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
using namespace std;
int n, m, tail, head, h[1000100], a[1000100];
int main()
{scanf("%d %d", &n, &m);tail = 0;head = 1;for (int i = 1; i <= n; ++i){scanf("%d", &a[i]);if (head <= tail && h[head] < i - m + 1) head++;//如果过了那就丢掉while(head <= tail && a[h[tail]] > a[i]) tail--;//如果比前面的小，那前面的丢掉h[++tail] = i;//存进去if (i >= m) printf("%d ", a[h[head]]);//这样下来最前面的就是最小的}tail = 0;head = 1;putchar(10);for (int i = 1; i <= n; ++i)//同上{if (head <= tail && h[head] < i - m + 1) head++;while(head <= tail && a[h[tail]] < a[i]) tail--;h[++tail] = i;if (i >= m) printf("%d ", a[h[head]]);}return 0;
}