codeforce723E One-Way Reform

题目链接
题意：给你一个无向图，不存在自环和重边，现在需要把图的无向边全部变成有向边，问最多可以使多少个结点的入度等于出度，输出所有有向边集合，不考虑边的顺序
思路： 容易联想到欧拉路，求解最多可以使多少个结点的入度等于出度很简单，统计无向图的偶数度数的结点个数即可，输出边却不怎么好写，一个好的思路，添加一个结点，连接上所有的奇数度数的结点，这样整个图就由若干个欧拉图组成，跑若干次欧拉图就可以了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 205;
const int maxm = maxn * maxn;
struct Edge
{
    int from,to,next;
};
Edge e[maxm];
int edgeNum,head[maxn],n,m,du[maxn];
bool vis[maxm];

void addEdge(int from,int to)
{
    e[edgeNum].from = from;
    e[edgeNum].to = to;
    e[edgeNum].next = head[from];
    head[from] = edgeNum;
    edgeNum++;
}

void dfs(int rt)
{
    for(int j = head[rt];j != -1; j = e[j].next)
    {
        if(vis[j]) continue;
        int to = e[j].to;
        if(rt && to) printf("%d %d\n",rt,to);
        vis[j] = vis[j^1] = true;
        dfs(to);
    }
}
int main()
{
#ifdef LOCAL_DEBUG
freopen("input.txt","r",stdin);
#endif // LOCAL_DEBUG
    int T;
    scanf("%d" ,&T);
    while(T--)
    {
        memset(head,-1,sizeof(head));
        memset(du,0,sizeof(du));
        memset(vis,0,sizeof(vis));
        scanf("%d%d" ,&n,&m);
        int from,to;
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d" ,&from,&to);
            addEdge(from,to);
            addEdge(to,from);
            du[from]++; du[to]++;
        }
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            if(du[i]%2 != 0)
            {
                addEdge(i,0); addEdge(0,i);
            }
            else sum++;
        }
        printf("%d\n",sum);
        for(int i = 0; i <= n; i++) dfs(i);
    }
    return 0;
}