欧拉回路必须满足每个点的度都是偶数。
字典序的话只用按照编号排序一以后从大到小建边。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;

const int N = 51, M = 2010;

int n, m, deg[N];
int tot, Head[N], ver[M << 1], Next[M << 1], id[M << 1]; bool v[M];
struct rode { int x, y, id; } a[M];
int siz, path[M];

void add(int x, int y, int z) {
    tot++;
    ver[tot] = y;
    id[tot] = z;
    Next[tot] = Head[x];
    Head[x] = tot;
}

bool cmp(rode p1, rode p2) { return p1.id > p2.id; }

void dfs(int x, int pre) {
    for (int i = Head[x]; i; i = Head[x]) {
        int y = ver[i]; Head[x] = Next[i];
        if (v[i >> 1]) continue;
        v[i >> 1] = 1;
        dfs(y, id[i]);
    }
    path[++siz] = pre;
}

int main() {
    while (1) {
        memset(deg, 0, sizeof(deg));
        tot = 1, memset(Head, 0, sizeof(Head));
        memset(v, 0, sizeof(v));
        siz = 0;
        int x, y, z, st; n = 0;
        for (m = 1;; m++) {
            scanf("%d%d", &x, &y);
            if (!x && !y) break;
            if (m == 1) st = min(x, y);
            n = max(n, max(x, y));
            scanf("%d", &z);
            a[m] = (rode){x, y, z};
            deg[x]++, deg[y]++;
        }
        m--;
        if (!m) break;
        bool bk = 1;
        for (int i = 1; i <= n; i++) 
            if (deg[i] & 1) { bk = 0; break; }
        if (!bk) { puts("Round trip does not exist."); continue; }
        sort(a + 1, a + m + 1, cmp);
        for (int i = 1; i <= m; i++) 
            add(a[i].x, a[i].y, a[i].id), add(a[i].y, a[i].x, a[i].id);
        dfs(st, 0); siz--;
        for (int i = siz; i > 1; i--) printf("%d ", path[i]);
        printf("%d\n", path[1]);
    }
    return 0;
}