51nod1437 迈克步 单调栈

考虑一个点作为最小值的区间$[L[i], R[i]]$

那么这个区间的所有含$i$的子区间最小值都是$v[i]$

因此，用单调栈求出$L[i], R[i]$后，对$R[i] - L[i] + 1$这个长度打一个$v[i]$的标记

之后，统计后缀最大值就能得出答案

注：不加输出优化会$T$

复杂度$O(n)$，暂居$rk1$

#include <cstdio>
#include <iostream>
using namespace std;

extern inline char gc() {
    static char RR[], *S = RR + , *T = RR + ;
    if(S == T) fread(RR, , , stdin), S = RR;
    return *S ++;
}
inline int read() {
    int p = , w = ; char c = gc();
    while(c > '' || c < '') { if(c == '-') w = -; p = p *  + c - ''; c = gc(); }
    while(c >= '' && c <= '') p = p *  + c - '', c = gc();
    return p * w;
}

int wr[], rw;
#define pc(o) *O ++ = o
char WR[], *O = WR;
inline void write(int x) {
    if(!x) pc('');
    if(x < ) x = -x, pc('-');
    while(x) wr[++ rw] = x % , x /= ;
    while(rw) pc(wr[rw --] + ''); pc(' ');
}

#define ri register int
#define sid 200050

int n, st[sid], top;
int v[sid], L[sid], R[sid], ans[sid];

int main() {
    n = read();
    for(ri i = ; i <= n; i ++) v[i] = read();

    st[top = ] = ; v[] = ;
    for(ri i = ; i <= n; i ++) {
        while(top && v[st[top]] >= v[i]) top --;
       L[i] = st[top] + ; st[++ top] = i;
    }

    st[top = ] = n + ; v[n + ] = ;
    for(ri i = n; i >= ; i --) {
        while(top && v[st[top]] >= v[i]) top --;
        R[i] = st[top] - ; st[++ top] = i;
    }

    for(ri i = ; i <= n; i ++) {
        int len = R[i] - L[i] + ;
        ans[len] = max(ans[len], v[i]);
    }
    for(ri i = n; i >= ; i --) ans[i] = max(ans[i], ans[i + ]);
    for(ri i = ; i <= n; i ++) write(ans[i]);
    fwrite(WR, , O - WR, stdout);
    return ;
}