与题意相反，从左往右进行构造。 注意到对每一段区间[L, R]选取最小对{x, y}时，x必须处于与L同奇偶性的位子，而y必须处于与x奇偶性相反的位子。选取最小的x以及最小的y可以用两个RMQ维护原数组，奇偶各一个。 当选完一对{x, y}时，区间分裂为[L, pos[x]-1], [pos[x]+1, pos[y]-1], [pos[y]+1, R]三个区间，可以用优先队列进行BFS，保证字典序最小。
代码：
#include<bits/stdc++.h> #define pb push_back #define fi first #define se second #define CLR(A, X) memset(A, X, sizeof(A)) using namespace std; typedef long long LL; typedef pair<int, int> PII; const double eps = 1e-10; int dcmp(double x) { if(fabs(x) < eps) return 0; return x>0?1:-1; } const LL MOD = 1e9+7; const int N = 2e5+5; const int INF = 0x3f3f3f3f; int d[2][N][32], a[2][N], pos[N]; void RMQ_init(int n, int v) { for(int i = 0; i < n; i++) d[v][i][0] = a[v][i]; for(int j = 1; (1<<j) <= n; j++) for(int i = 0; i+(1<<j)-1 < n; i++) d[v][i][j] = min(d[v][i][j-1], d[v][i+(1<<(j-1))][j-1]); } int RMQ(int L, int R, int v) { int k = 0; while((1<<(k+1)) <= R-L+1) k++; return min(d[v][L][k], d[v][R-(1<<k)+1][k]); } struct node { int x, L, R; bool operator < (const node& A) const { return x > A.x; } }; int main() { int n, x; scanf("%d", &n); for(int i = 0; i < n; i++) { scanf("%d", &x); a[i&1][i] = x; a[(i&1)^1][i] = INF; pos[x] = i; } RMQ_init(n, 0); RMQ_init(n, 1); priority_queue<node> Q; x = RMQ(0, n-1, 0); Q.push({x, 0, n-1}); while(!Q.empty()) { node x = Q.top(); Q.pop(); int v1 = x.x, v2 = RMQ(pos[x.x]+1, x.R, (pos[x.x]&1)^1); printf("%d %d ", v1, v2); v1 = pos[v1], v2 = pos[v2]; if(x.L != v1) Q.push({RMQ(x.L, v1-1, x.L&1), x.L, v1-1}); if(v1+1 != v2) Q.push({RMQ(v1+1, v2-1, (v1+1)&1), v1+1, v2-1}); if(v2 != x.R) Q.push({RMQ(v2+1, x.R, (v2+1)&1), v2+1, x.R}); } return 0; }