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Note that girls in Arpa’s land are really attractive.

Arpa loves overnight parties. In the middle of one of these parties Mehrdad suddenly appeared. He saw n pairs of friends sitting around a table. i-th pair consisted of a boy, sitting on the ai-th chair, and his girlfriend, sitting on the bi-th chair. The chairs were numbered 1through 2n in clockwise direction. There was exactly one person sitting on each chair.

There were two types of food: Kooft and Zahre-mar. Now Mehrdad wonders, was there any way to serve food for the guests such that: 

• Each person had exactly one type of food, 
• No boy had the same type of food as his girlfriend, 
• Among any three guests sitting on consecutive chairs, there was two of them who had different type of food. Note that chairs 2n and 1 are considered consecutive. 

Find the answer for the Mehrdad question. If it was possible, find some arrangement of food types that satisfies the conditions.

31 42 53 6

1 22 11 2

题意：

给出n，表示有2n个人围成一圈坐在桌子边上，对应这2n个人是n对情侣，每个人占据一个位子，然后n行，每行两个数，表示坐在这两个位置的人是情侣，要求情侣不能吃同一种食物，并且桌子上相邻的三个人的食物必须有两个人是不同的，只有两种食物（1或者是2），问一种可行分配方式。不能分配输出-1.

题解：

一开始看别人的题解，都是说情侣之间连边，2*i和2*i-1位置的人连边，然后进行二分染色，我照这个思路写了代码，然后A了，不过不知道为什么是2*i和2*i-1之间连边。

其实，这样的连边一定能保证图是个二分图，即一定有可行分配方式，因为图中不存在长度为奇数的环，所以是个二分图，而且恰好这样连边也满足了题意。

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define rep(i,a,n) for (int i=a;i
          
           =a;i--)#define pb push_back#define fi first#define se secondtypedef vector
           
             VI;typedef long long ll;typedef pair
            
              PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=2e5+100;int head[maxn];struct edge{ int to,next;}e[maxn*10]; //int tol=0;void add(int u,int v){ e[++tol].to=v,e[tol].next=head[u],head[u]=tol;}int c[maxn];bool dfs(int u,int v){ c[u]=v; for(int i=head[u];i;i=e[i].next) { int vv=e[i].to; if(c[vv]&&c[vv]==v) return false; else if(!c[vv]&&!dfs(vv,3-v)) return false; } return true;}vector
             
               V;int main(){ int n; scanf("%d",&n); rep(i,1,n+1) { int u,v; scanf("%d%d",&u,&v); add(u,v),add(v,u); V.pb(make_pair(u,v)); } for(int i=2;i<=2*n;i+=2) { add(i,i-1),add(i-1,i); } memset(c,0,sizeof(c)); rep(i,1,2*n+1) { if(!c[i]) { if(!dfs(i,1)) { puts("-1"); return 0; } } } rep(i,0,n) { int u=V[i].first,v=V[i].second; printf("%d %d\n",c[u],c[v]); } return 0;}
             
            
           
          
         
        
       
      
     
    
   

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