codeforces 547B. Mike and Feet 单调栈

题目链接

用单调栈计算出一个数字， 左边第一个比他小的数字的位置， 右边比第一个他小的数字的位置， 然后len = r[i] - l[i] +1. ans[len] = max(ans[len], a[i])

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+5;
int l[maxn], r[maxn], ans[maxn], a[maxn], q[maxn];
int main()
{
    int n;
    cin>>n;
    for(int i = 1; i<=n; i++) {
        scanf("%d", &a[i]);
    }
    int i = 1, st = 0;
    while(i<=n) {
        while(st&&a[q[st-1]]>a[i]) {
            r[q[st-1]] = i-1;
            st--;
        }
        q[st++] = i;
        i++;
    }
    while(st) {
        r[q[st-1]] = n;
        st--;
    }
    i = n;
    while(i>=1) {
        while(st&&a[q[st-1]]>a[i]) {
            l[q[st-1]] = i+1;
            st--;
        }
        q[st++] = i;
        i--;
    }
    while(st) {
        l[q[st-1]] = 1;
        st--;
    }
    memset(ans, 0, sizeof(ans));
    for(int i = 1; i<=n; i++) {
        int len = r[i]-l[i]+1;
        if(ans[len]<a[i]) {
            ans[len] = a[i];
        }
    }
    for(i = n; i>=1; i--) {
        if(ans[i]>=ans[i-1])
            ans[i-1] = ans[i];
    }
    for(int i = 1; i<=n; i++) {
        printf("%d ", ans[i]);
    }
}