枚举每个点作为次大值,用单调栈处理出左边 / 右边第一个比它大的数,直接回答
复杂度$O(n)$
#include <cstdio> #include <cstring> #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; c = gc(); } while(c >= ‘0‘ && c <= ‘9‘) p = p * 10 + c - ‘0‘, c = gc(); return p * w; } #define sid 200050 #define ri register int int n, top, ans; int st[sid], s[sid], L[sid], R[sid]; int main() { n = read(); for(ri i = 1; i <= n; i ++) s[i] = read(); st[top = 1] = 0; s[0] = 1e9 + 5; for(ri i = 1; i <= n; i ++) { while(top && s[st[top]] <= s[i]) top --; L[i] = st[top]; st[++ top] = i; } st[top = 1] = n + 1; s[n + 1] = 1e9 + 5; for(ri i = n; i >= 1; i --) { while(top && s[st[top]] <= s[i]) top --; R[i] = st[top]; st[++ top] = i; } for(ri i = 1; i <= n; i ++) { if(L[i] != 0) ans = max(ans, s[i] ^ s[L[i]]); if(R[i] != n + 1) ans = max(ans, s[i] ^ s[R[i]]); } printf("%d\n", ans); return 0; }
原文地址:https://www.cnblogs.com/reverymoon/p/9557993.html
时间: 2024-10-03 04:05:31