CodeForces 527E Data Center Drama（欧拉回路）_加边变成欧拉路 codeforces

题目链接：codeforces.com/problemset/problem/527/E

题意：给出一个无向联通图，把无向边变成有向边，并添加最少的有向边使得所有结点入度出度都是偶数

思路：最终构造出来的图是一个欧拉回路，对于一张无向图，当其所有点度数为偶数时存在欧拉回路，如果回路边数为偶数则一定能构造出来，所以对于一开始度数是奇数的点依次添加无向边使它们度数为偶数，然后对于这张无向图，任选起点跑欧拉回路，构造类似a - > b <- c -> d...的路径，如果跑出的路径为奇数任选一点添加自环即可，注意要将之前访问过的边拆掉以防超时

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>
 
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")
 
using namespace std;
 
const int maxn = 500100;
const int inf = 0x3f3f3f3f;
 
int head[maxn], cnt;
 
int deg[maxn], vis[maxn];
vector <int> ans;
 
struct edge
{
    int from, to, nxt;
} e[maxn];
 
void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
    memset(deg, 0, sizeof(deg));
    memset(vis, 0, sizeof(vis));
}
 
void add(int u, int v)
{
    e[cnt].from = u;
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt++;
}
 
void dfs(int u)
{
    for (int &i = head[u]; ~i; i = e[i].nxt)
    {
        if (vis[i / 2]) continue;
        vis[i / 2] = 1;
        int v = e[i].to;
        int tmp = i;
        dfs(v);
        ans.push_back(tmp);
    }
}
 
int main()
{
    int n, m;
    while (~scanf("%d%d", &n, &m))
    {
        init();
        while (m--)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            add(u, v);
            add(v, u);
            deg[u]++, deg[v]++;
        }
 
        int p = -1;
        for (int i = 1; i <= n; i++)
        {
            if (deg[i] & 1)
            {
                if (p == -1)
                    p = i;
                else
                {
                    add(i, p);
                    add(p, i);
                    deg[i]++, deg[p]++;
                    p = -1;
                }
            }
        }
        if ((cnt / 2) & 1) add(n, n);   //加自环
 
        ans.clear();
        dfs(1);
        cout << ans.size() << endl;
        for (int i = 0; i < ans.size(); i++)
        {
            if (i & 1)
                printf("%d %d\n", e[ans[i]].from, e[ans[i]].to);
            else
                printf("%d %d\n", e[ans[i]].to, e[ans[i]].from);
        }
    }
    return 0;
}