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[23456], *S = RR + 23333, *T = RR + 23333; if(S == T) fread(RR, 1, 23333, stdin), S = RR; return *S ++; } inline int read() { int p = 0, w = 1; char c = gc(); while(c > '9' || c < '0') { if(c == '-') w = -1; p = p * 10 + c - '0'; c = gc(); } while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc(); return p * w; } int wr[50], rw; #define pc(o) *O ++ = o char WR[30000005], *O = WR; inline void write(int x) { if(!x) pc('0'); if(x < 0) x = -x, pc('-'); while(x) wr[++ rw] = x % 10, x /= 10; while(rw) pc(wr[rw --] + '0'); 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 = 1; i <= n; i ++) v[i] = read(); st[top = 1] = 0; v[0] = 0; for(ri i = 1; i <= n; i ++) { while(top && v[st[top]] >= v[i]) top --; L[i] = st[top] + 1; st[++ top] = i; } st[top = 1] = n + 1; v[n + 1] = 0; for(ri i = n; i >= 1; i --) { while(top && v[st[top]] >= v[i]) top --; R[i] = st[top] - 1; st[++ top] = i; } for(ri i = 1; i <= n; i ++) { int len = R[i] - L[i] + 1; ans[len] = max(ans[len], v[i]); } for(ri i = n; i >= 1; i --) ans[i] = max(ans[i], ans[i + 1]); for(ri i = 1; i <= n; i ++) write(ans[i]); fwrite(WR, 1, O - WR, stdout); return 0; }