欧拉回路是简单回路_欧拉回路的充分必要条件

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

using namespace std;

//———————————————-

typedef pair 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

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

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 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 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;

}