单调栈

1437 迈克步
一个长度的值为所有该长度区间最小值中的最大值,问你所有的长度的这个值.
用单调栈,得出每个是最小值的最左和最右,就知道他的贡献范围了.
最后线性处理一下.

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <set>
using namespace std;
#ifdef noob
#define debug(x) std::cerr << #x << " = " << (x) << std::endl
#else
#define debug(...)
#endif
typedef long long LL;
const int MAXN = 2e5 + 17;
const int MOD = 1e9 + 7;
int a[MAXN],lf[MAXN],rg[MAXN],mx[MAXN];
int main()
{
#ifdef noob
    freopen("Input.txt","r", stdin);
    freopen("Output.txt", "w", stdout);
#endif
    int n,sz=1;
    cin>>n;
    for (int i = 0; i < n; ++i)
        scanf("%d",a+i);
    vector<pair<int,int > > vec;
    vec.push_back({0,a[0]});
    for (int i = 1; i < n; ++i)
    {
        while(sz>0&&a[i]<vec[sz-1].second)
        {
            rg[vec[sz-1].first] = i;
            vec.pop_back();
            sz--;
        }
        vec.push_back({i,a[i]});
        sz++;
    }
    for (int i = 0; i < sz; ++i) rg[vec[i].first] = n;
    vec.clear();
    sz = 1;
    vec.push_back({n-1,a[n-1]});
    for (int i = n-2; i > -1; --i)
    {
        while(sz>0&&a[i]<vec[sz-1].second)
        {
            lf[vec[sz-1].first] = i;
            vec.pop_back();
            sz--;
        }
        vec.push_back({i,a[i]});
        sz++;
    }
    for (int i = 0; i < sz; ++i) lf[vec[i].first] = -1;
    vector<pair<int,int > > zt;
    for (int i = 0; i < n; ++i)
        zt.push_back({a[i],rg[i]-lf[i]});
    sort(zt.begin(), zt.end());
    int st = 0;
    for (int i = n-1; i > -1; --i)
    {
        for (int j = st; j < zt[i].second; ++j)
            mx[j] = zt[i].first;
        st = max(zt[i].second,st);
    }
    for (int i = 0; i < n; ++i)
        printf("%d%c",mx[i+1],i==n-1?'\n':' ' );
    return 0;
}