Codeforces 723E One-Way Reform(歐拉回路)

給你n點m邊的圖，然後讓你確定每條邊的方向，使得入度=出度的點最多
猜想所有偶數度數的點都能做到入度=出度
如何確定方向呢，考慮到裏面奇數度數的點一定是偶數個
假設他們是v1,v2....,v2k
把v1與v2,v3與v4....v2k−1與v2k連邊
這樣所有點都是偶數度數，對於每個連通塊存在歐拉回路
然後從每個點開始dfs就行了，走過的邊就記錄方向，然後標記不能走
碰到是新加的邊，就不放進答案

---

代碼：

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#pragma comment(linker,"/STACK:102400000,102400000")

using namespace std;
#define   MAX           210
#define   MAXN          1000005
#define   maxnode       5
#define   sigma_size    30
#define   lson          l,m,rt<<1
#define   rson          m+1,r,rt<<1|1
#define   lrt           rt<<1
#define   rrt           rt<<1|1
#define   middle        int m=(r+l)>>1
#define   LL            long long
#define   ull           unsigned long long
#define   mem(x,v)      memset(x,v,sizeof(x))
#define   lowbit(x)     (x&-x)
#define   pii           pair<int,int>
#define   bits(a)       __builtin_popcount(a)
#define   mk            make_pair
#define   limit         10000

//const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const LL     INFF  = 0x3f3f;
//const double PI    = acos(-1.0);
const double inf   = 1e18;
const double eps   = 1e-6;
const LL     mod   = 1e9+7;
const ull    mx    = 133333331;

/*****************************************************/
inline void RI(int &x) {
      char c;
      while((c=getchar())<'0' || c>'9');
      x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
 }
/*****************************************************/

struct Edge{
    int v,next,c;
}edge[MAX*MAX*2];
int head[MAX];
int deg[MAX];
int tot;
vector<pii> ans;
void init(){
    mem(head,-1);
    mem(deg,0);
    tot=0;
}

void add_edge(int a,int b,int c){
    edge[tot]=(Edge){b,head[a],c};
    head[a]=tot++;
}

void dfs(int u){
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].v;
        if(edge[i].c==-1) continue;
        if(edge[i].c==0) ans.push_back(mk(u,v));
        edge[i].c=edge[i^1].c=-1;
        dfs(v);
    }
}
int main(){
    //freopen("in.txt","r",stdin);
    int t;
    cin>>t;
    while(t--){
        int n,m;
        cin>>n>>m;
        init();
        for(int i=0;i<m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            add_edge(a,b,0);
            add_edge(b,a,0);
            deg[a]++;
            deg[b]++;
        }
        vector<int> v;
        int cnt=0;
        for(int i=1;i<=n;i++){
            if(deg[i]%2) v.push_back(i);
            else cnt++;
        }
        for(int i=0;i<v.size();i+=2){
            int a=v[i];
            int b=v[i+1];
            add_edge(a,b,1);
            add_edge(b,a,1);
        }
        ans.clear();
        for(int i=1;i<=n;i++) dfs(i);
        cout<<cnt<<endl;
        for(int i=0;i<ans.size();i++){
            cout<<ans[i].first<<" "<<ans[i].second<<endl;
        }
    }
    return 0;
}