164 lines
4.7 KiB
C++
164 lines
4.7 KiB
C++
/*
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https://codeforces.com/problemset/problem/36/E
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欧拉路+并查集+思路~
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简直要被绕晕了……有好多细节要注意……
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首先，要分为只有一个联通块，有一个联通块和有两个及以上联通块，判断联通块数用并查集。
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最后一种是不行的，直接输出-1。
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对于其余两种，要判断每个点的入度是否满足欧拉路条件（没有奇点或有两个奇点），特别的，对于第二种，
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每个连通块内要分别判断。
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然后，如果有两个奇点，从其中一个出发遍历即可。
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在第一种情况中，如果有0个或2个起点，则求出总的欧拉路后从任意处截断。
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如果有4个起点，任选两个连边后求欧拉路，再在连的边处截断。
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https://blog.csdn.net/xianpingping/article/details/82468730
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*/
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#include <cstdio>
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#include <cstring>
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#include <iostream>
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#include <algorithm>
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#define M 10100
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using namespace std;
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struct abcd {
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int to, num, next;
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bool flag;
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} table[M << 1];
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int head[M], tot = 1;
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int n = 10000, m, a[2], degree[M];
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int stack[M], top;
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void Add(int x, int y, int z) {
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degree[y]++;
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table[++tot].to = y;
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table[tot].num = z;
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table[tot].next = head[x];
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head[x] = tot;
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}
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namespace Union_Find_Set {
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int fa[M], rank[M];
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int Find(int x) {
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if (!fa[x] || fa[x] == x)
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return fa[x] = x;
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return fa[x] = Find(fa[x]);
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}
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void Union(int x, int y) {
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x = Find(x);
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y = Find(y);
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if (x == y) return;
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if (rank[x] > rank[y])
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swap(x, y);
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if (rank[x] == rank[y])
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rank[y]++;
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fa[x] = y;
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}
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} // namespace Union_Find_Set
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void DFS(int x) {
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int i;
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for (i = head[x]; i; i = head[x]) {
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head[x] = table[head[x]].next;
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if (table[i].flag)
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continue;
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table[i ^ 1].flag = true;
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DFS(table[i].to);
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stack[++top] = table[i].num;
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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("CF36E.in", "r", stdin);
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#endif
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using namespace Union_Find_Set;
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int i, x, y;
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cin >> m;
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if (m == 1) {
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cout << -1 << endl;
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return 0;
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}
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for (i = 1; i <= m; i++) {
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scanf("%d%d", &x, &y);
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Add(x, y, i);
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Add(y, x, i);
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Union(x, y);
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}
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for (i = 1; i <= n; i++)
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if (head[i]) {
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if (!a[0])
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a[0] = i;
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else if (Find(i) != Find(a[0])) {
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if (!a[1])
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a[1] = i;
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else if (Find(i) != Find(a[1]))
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return cout << -1 << endl, 0;
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}
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}
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if (a[1]) {
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int cnt[2] = {0, 0};
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int st[2] = {a[0], a[1]};
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for (i = 1; i <= n; i++)
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if (degree[i] & 1) {
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cnt[Find(i) == Find(a[1])]++;
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st[Find(i) == Find(a[1])] = i;
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}
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if (cnt[0] > 2 || cnt[1] > 2)
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return cout << -1 << endl, 0;
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DFS(st[0]);
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cout << top << endl;
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while (top)
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printf("%d%c", stack[top], top == 1 ? '\n' : ' '), top--;
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DFS(st[1]);
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cout << top << endl;
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while (top)
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printf("%d%c", stack[top], top == 1 ? '\n' : ' '), top--;
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} else {
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int st[4] = {0, 0, 0, 0};
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for (i = 1; i <= n; i++)
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if (degree[i] & 1) {
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if (!st[0])
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st[0] = i;
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else if (!st[1])
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st[1] = i;
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else if (!st[2])
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st[2] = i;
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else if (!st[3])
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st[3] = i;
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else
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return cout << -1 << endl, 0;
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}
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if (!st[0]) {
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DFS(a[0]);
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cout << 1 << '\n'
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<< stack[top--] << endl;
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cout << top << endl;
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while (top)
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printf("%d%c", stack[top], top == 1 ? '\n' : ' '), top--;
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} else if (!st[2]) {
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DFS(st[0]);
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cout << 1 << '\n'
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<< stack[top--] << endl;
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cout << top << endl;
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while (top)
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printf("%d%c", stack[top], top == 1 ? '\n' : ' '), top--;
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} else {
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Add(st[0], st[1], m + 1);
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Add(st[1], st[0], m + 1);
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DFS(st[2]);
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for (i = 1; i <= top; i++)
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if (stack[i] == m + 1)
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break;
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cout << top - i << endl;
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while (top > i)
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printf("%d%c", stack[top], top == i + 1 ? '\n' : ' '), top--;
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top--;
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cout << top << endl;
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while (top)
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printf("%d%c", stack[top], top == 1 ? '\n' : ' '), top--;
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}
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}
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return 0;
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} |