CodeForces 817D : Mike and Feet 单调栈

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传送门

题意

n个值代表n个熊的高度对于size为x的group strength值为这个group中熊的最小的height值

对于x（1<=x<=n）求出最大的strength值

分析

首先我们知道，大范围的答案可以往小范围转移
我们用单调栈处理出来每个数的区间，然后对应最大的答案，最后从大到小递推一下答案即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e6 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int l[N], r[N]; 
int s[N];
int t;
int a[N];
int n;
int ans[N];int main() {read(n);for (int i = 1; i <= n; i++) read(a[i]);t = 0;s[t] = 0;for (int i = 1; i <= n; i++) {while (t && a[s[t]] >= a[i]) t--;l[i] = s[t] + 1;s[++t] = i;}t = 0;s[t] = n + 1;for (int i = n; i; i--) {while (t && a[s[t]] >= a[i]) t--;r[i] = s[t] - 1;s[++t] = i;}for(int i = 1;i <= n;i++){int x = r[i] - l[i] + 1;ans[x] = max(ans[x],a[i]);}  int mx = 0;for(int i = n;i;i--){ans[i] = max(ans[i],mx);mx = max(ans[i],mx);}for(int i = 1;i <= n;i++) printf("%d ",ans[i]);return 0;
}