【2017.12.22.A】

                            A

题面：

给一个n个点m条边的无向图，你可以选择一个点作为起点，然后沿着图中的边开始走，走的过程中，同一条边不能经过两次（相反的方向也不行）。最后你要回到起点。假设你走了x条边，有两个问题：

Q1：x 是否可以为奇数？

Q2：x 是否可以为偶数？

输入：

第一行读入两个数 n 和 m，分别表示图的点数和边数。

接下来m行，每行两个整数ui, vi，表示ui和vi之间有一条边。

保证不会有自环和重边。

输出：

输出两行，第一行输出YES或NO，表示第一个问题的答案，第二行输出YES或NO，表示第二个问题的答案。

样例输入：

1 0

样例输出：

NO

NO

样例输入：

3 3

1 2

2 3

3 1

样例输出：

YES

NO

样例输入：

4 4

1 2

2 3

3 4

4 1

样例输出：

NO

YES

数据范围：

对于20%的数据，1 <= n <= 30, 1 <= m <= 50

对于50%的数据，1 <= n <= 1000, 1 <= m <= 3000

对于100%的数据，1 <= n <= 100000, 1 <= m <= 300000, 1 <= ui, vi <= n

【题解】

        ①对于我们的图走啊走你会发现这样一件事：

          若一个点只在一个奇环或偶环中，那么从它出发必定只能走出奇或偶；

          若一个点只在多个奇环中 或者 在奇环和偶环中，那么从它出发奇偶都可以走；

          若一个点只在多个偶环中，那么从它出发只能走出偶；（奇偶都指边数）

        ②考虑记下每个点被多少个奇环和偶环覆盖dfs+差分即可；（如果不是很清楚可以百度树上差分）

          注意差分的技巧性（因标记边，点而异，此处为标记点）：

        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" 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 /*5 6
 1 2
 2 3
 1 3
 3 4
 4 5
 3 5
 在下面自己测wa得一塌糊涂，交到cena就AC了；
 我还搞了两个多小时；
 只要有有漏洞的C++在，就没有上帝。。
 记住树上差分的基本思想。（不过我现在还不知道染色到底有没有理）
 */
 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #include <queue>
 #include <vector>
 #include <ctime>
 #include <cmath>
 #define inf 0x3f3f3f3f
 #define ll long long
 #define N 100100
 #define M 300100
 #define mem(f,a) memset(f,a,sizeof(f))
 #define Run(i,l,r) for(int i=l;i<=r;i++)
 #define Don(i,l,r) for(int i=l;i>=r;i--)
 #define Eun(i,u,hd,E) for(int i=hd[u],v=E[i].v;i!=-1;i=E[i].next,v=E[i].v)
 using namespace std;
 struct Edge{
     int u,v,next;
 }E[*M],e[M];
 int n,m;
 int k1,k2,hd1[N],hd2[N],fg1,fg2;
 int c1[N],c2[N],p[N][],d[N],vis[*M],idx,f[N];
 void adde1(int u,int v)
 {    E[k1]=(Edge){u,v,hd1[u]};
     hd1[u]=k1++;
 }
 void adde2(int u,int v)
 {    e[k2]=(Edge){u,v,hd2[u]};
     hd2[u]=k2++;
 }
 int find(int x)
 {    if (f[x]==x) return x;
     return f[x]=find(f[x]);
 }
 void Dfs(int u,int fa)
 {    p[u][]=fa; d[u]=d[fa]+;
     Run(i,,){
         if ((<<i)>d[u]) break;
         p[u][i]=p[p[u][i-]][i-];
     }
     Eun(i,u,hd2,e) Dfs(v,u);
 }
 int LCA(int a,int b)
 {    if (d[a]>d[b]) swap(a,b);
     Don(i,,)    if (d[a]<=d[b]-(<<i)) b=p[b][i];
     if (a==b) return a;
     Don(i,,) if (p[a][i]!=p[b][i]) a=p[a][i],b=p[b][i];
     return p[a][];
 }
 void update(int u,int v)
 {    int t=LCA(u,v);
     if ((d[v]+d[u]-*d[t]+)&) c1[t]--,c1[p[t][]]--,c1[u]++,c1[v]++;
     else c2[t]--,c2[p[t][]]--,c2[u]++,c2[v]++;
 }
 void dfs1(int u,int fa)
 {    int fu=find(u);
     Eun(i,u,hd1,E){
         int fv=find(v);
         if (v==fa) vis[i]=;
         if (fu==fv) continue;
         vis[i]=;adde2(u,v);f[fv]=fu;
         dfs1(v,u);
     }
 }
 void dfs2(int u)
 {    Eun(i,u,hd2,e) dfs2(v),c1[u]+=c1[v],c2[u]+=c2[v];
     fg1|=c1[u]; fg2|=c2[u];
     if (c1[u]>) fg2=;
 }
 int main()
 {    freopen("A.in","r",stdin);
     //freopen("A.out","w",stdout);
     scanf("%d%d",&n,&m);
     mem(hd1,-); mem(hd2,-);
     Run(i,,m){
         int u,v;
         scanf("%d%d",&u,&v);
         adde1(u,v);adde1(v,u);
     }
     Run(i,,n) f[i]=i;
     Run(i,,n) if (f[i]==i) dfs1(i,);    //建树
     Run(i,,n) if (f[i]==i) Dfs(i,);     //LCA预处理
     Run(i,,*m-)  if (!vis[i]) update(E[i].u,E[i].v),vis[i^]=;
     Run(i,,n) if (f[i]==i) dfs2(i);      //差分
     if (fg1) printf("YES\n"); else printf("NO\n");
     if (fg2) printf("YES\n"); else printf("NO\n");
     return ;
 }//by tkys_Austin;