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

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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;}