UOJ117. 欧拉回路【欧拉回路模板题】

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题目大意

就是让你对有向图和无向图分别求欧拉回路

非常的模板，但是由于UOJ上毒瘤群众太多了

所以你必须加上一个小优化

就是每次访问过一个边就把它删掉

有点像Dinic的当前弧优化的感觉

注意是在dfs完一个节点把当前的边加入到栈里面

然后输出的时候为了保证原来的顺序就直接弹栈就好了

---

//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
typedef pair<int, int> pi;
typedef long long ll;
typedef double db;
#define fi first
#define se second
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
template <typename T>
void Read(T &x) {
  bool w = 1;x = 0;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') w = 0, c = getchar();
  while (isdigit(c)) {
    x = (x<<1) + (x<<3) + c -'0';
    c = getchar();
  }
  if (!w) x = -x;
}
template <typename T>
void Write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) Write(x / 10);
  putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 4e5 + 10;
struct Edge {
  int v, id, nxt;
} E[N];
int head[N], tot = 0;
int fro[N], to[N], n, m;
void addedge(int u, int v, int id) {
  E[++tot] = (Edge) {v, id, head[u]};
  head[u] = tot;
}
namespace Solve1 {
int du[N], vis[N];
stack<int> st;
void dfs(int u) {
  for (int &i = head[u]; i; i = E[i].nxt) {
    int cur = i;
    if (vis[abs(E[cur].id)]) continue;
    vis[abs(E[cur].id)] = 1;
    dfs(E[cur].v);
    st.push(E[cur].id);
  }
}
void solve() {
  fu(i, 1, m) ++du[fro[i]], ++du[to[i]];
  fu(i, 1, n) if (du[i] & 1) {
    printf("NO");
    return;
  }
  fu(i, 1, m) {
    addedge(fro[i], to[i], i);
    addedge(to[i], fro[i], -i);
  }
  fd(i, n, 1) {
    if (head[i]) {
      dfs(i);
      break;
    }
  }
  if ((signed) st.size() != m) {
    printf("NO");
  } else {
    printf("YES\n");
    while (st.size()) {
      Write(st.top()), putchar(' ');
      st.pop();
    }
  }
}
}
namespace Solve2 {
int in[N], out[N], vis[N];
stack<int> st;
void dfs(int u) {
  for (int &i = head[u]; i; i = E[i].nxt) {
    int cur = i;
    if (vis[E[cur].id]) continue;
    vis[E[cur].id] = 1;
    dfs(E[cur].v);
    st.push(E[cur].id);
  }
}
void solve() {
  fu(i, 1, m) ++out[fro[i]], ++in[to[i]];
  fu(i, 1, n) if (out[i] ^ in[i]) {
    printf("NO");
    return;
  }
  fu(i, 1, m) addedge(fro[i], to[i], i);
  fu(i, 1, n) {
    if (head[i]) {
      dfs(i);
      break;
    }
  }
  if ((signed) st.size() != m) {
    printf("NO");
  } else {
    printf("YES\n");
    while (st.size()) {
      Write(st.top()), putchar(' ');
      st.pop();
    }
  }
}
}
int main() {
#ifdef dream_maker
  freopen("input.txt", "r", stdin);
#endif
  int op; Read(op);
  Read(n), Read(m);
  fu(i, 1, m) Read(fro[i]), Read(to[i]);
  if (op == 1) Solve1::solve();
  else Solve2::solve();
  return 0;
}