Codeforces 723E One-Way Reform(欧拉回路)

给你n点m边的图，然后让你确定每条边的方向，使得入度=出度的点最多

猜想所有偶数度数的点都能做到入度=出度

如何确定方向呢，考虑到里面奇数度数的点一定是偶数个

假设他们是v1,v2....,v2k

把v1与v2,v3与v4....v2k−1与v2k连边

这样所有点都是偶数度数，对于每个连通块存在欧拉回路

然后从每个点开始dfs就行了，走过的边就记录方向，然后标记不能走

碰到是新加的边，就不放进答案

代码：

#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;
}