Macaron_lin @ 2019-04-28 17:08:04
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 8000000 + 10;
int n, m;
int num[maxn];
struct Tr
{
int l, r;
int v;
}tr[maxn];
void Read()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> num[i];
}
}
void Build(int L, int R, int i)
{
tr[i].l = L; tr[i].r = R;//区间范围
if (L == R)
{
tr[i].v = num[L];
}
else
{
int M = (L + R) / 2;
Build(L, M, 2 * i);
Build(M + 1, R, 2 * i + 1);
tr[i].v = min(tr[2 * i].v, tr[2 * i + 1].v);
}
}
int Fun(int qL, int qR, int i)
{
int L = tr[i].l, R = tr[i].r;
if (qL <= L && R <= qR) return tr[i].v;
int ans = 1e8, M = (L + R) / 2;
if (qL <= M) ans = min(ans, Fun(qL, qR, 2 * i));
if (M + 1 <= qR) ans = min(ans, Fun(qL, qR, 2 * i + 1));
return ans;
}
void Print()
{
for (int i = 1; i <= n; i++)
{
if (i == 1) cout << 0 << '\n';
else cout << Fun(max(1, i - m), i - 1, 1) << '\n';
}
}
int main()
{
ios::sync_with_stdio(false);
Read();
Build(1, n, 1);
Print();
getchar(); getchar();
return 0;
}
求助啊(委屈
by yummy @ 2019-04-28 17:18:44
正解不是单调队列吗
单调队列O(N),线段树O(NlogN)
by ecnerwaIa @ 2019-04-28 17:21:37
@嘿小恐龙 可以试着手写read,然后Fun函数改一下,可以卡卡常
by ecnerwaIa @ 2019-04-28 17:25:37
能用位运算肯定位运算啊
by MSwalker @ 2019-04-28 17:28:36
build里边没return啊
by wxwoo @ 2019-04-28 17:39:12
您最后的那两个getchar()
不应该删掉吗......
by VenusM1nT @ 2019-04-28 17:47:56
理性分析一下,线段树每次查询
by VenusM1nT @ 2019-04-28 17:52:10
啊说错了【我是智障】
by Smile_Cindy @ 2019-04-28 17:52:10
@Venus
确认线段树每次查询O(nlogn)???
by Smile_Cindy @ 2019-04-28 17:52:26
我可能学了假的线段树
by Juan_feng @ 2019-04-28 17:53:19
神仙Venus写错了吧/kk