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;
}