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#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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#include <cmath>
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#include <stack>
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#include <queue>
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#include <vector>
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#include <map>
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#include <string>
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#include <iostream>
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using namespace std;
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const int N = 2010, M = N * N;
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// 声明结构体
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struct Node {
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int a, b, c;
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bool const operator<(const Node &t) const {
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return c > t.c;
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}
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} g[N];
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int st[M]; // 某条边是不是访问过，无向图一般用于处理成对就换的边
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int d[N]; // 度
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int res[M], rl; // 路径数组
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int m; // m条边
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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void dfs(int u) {
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for (int i = h[u]; ~i; i = h[u]) { // 删边优化
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h[u] = ne[i];
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if (st[i]) continue;
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st[i] = st[i ^ 1] = 1; // 成对变换
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dfs(e[i]);
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// 记录路径
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res[++rl] = w[i];
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}
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}
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("POJ1041.in", "r", stdin);
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#endif
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int a, b, c;
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while (true) {
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memset(d, 0, sizeof d);
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memset(st, 0, sizeof st);
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memset(h, -1, sizeof h);
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memset(res, 0, sizeof res);
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idx = rl = m = 0;
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// 本题的输入挺奇怪的，需要使用while(true)+do while的方式来读取最方便
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scanf("%d%d", &a, &b);
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if (a == 0 && b == 0) exit(0);
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do {
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scanf("%d", &c);
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g[m].a = a, g[m].b = b, g[m++].c = c;
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scanf("%d%d", &a, &b);
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} while (a && b);
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// 针对边的边号进行由小到大排序
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sort(g, g + m);
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b, c = g[i].c;
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add(a, b, c), add(b, a, c);
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d[a]++, d[b]++;
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}
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int flag = 0;
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for (int i = 0; i < m; i++) {
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int a = g[i].a, b = g[i].b;
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if ((d[a] & 1) || (d[b] & 1)) {
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flag = 1;
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break;
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}
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}
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if (!flag) {
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dfs(1);
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// 逆序输出序列
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for (int i = rl; i; i--) printf("%d ", res[i]);
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} else
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cout << "Round trip does not exist.";
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puts("");
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}
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return 0;
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} |
