C. Graph Reconstruction

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/329/problem/C

Description

I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. For each pair of nodes, there is at most an edge connecting them. No edge connects a node to itself.

I would like to create a new graph in such a way that:

The new graph consists of the same number of nodes and edges as the old graph.

The properties in the first paragraph still hold.

For each two nodes u and v, if there is an edge connecting them in the old graph, there is no edge connecting them in the new graph.

Help me construct the new graph, or tell me if it is impossible.

Input

The first line consists of two space-separated integers: n and m (1 ≤ m ≤ n ≤ 10⁵), denoting the number of nodes and edges, respectively. Then m lines follow. Each of the m lines consists of two space-separated integers u and v (1 ≤ u, v ≤ n; u ≠ v), denoting an edge between nodes u and v.

Output

If it is not possible to construct a new graph with the mentioned properties, output a single line consisting of -1. Otherwise, output exactly m lines. Each line should contain a description of edge in the same way as used in the input format.

Sample Input

8 7 1 2 2 3 4 5 5 6 6 8 8 7 7 4

Sample Output

1 4 4 6 1 6 2 7 7 5 8 5 2 8

HINT

题意

给你一个n个点m条边的无向图

无自环，无重边，每个点的度数最多为2

然后让你找到一个图，使得性质一样，但是之前相连的边，之后不能相连

题解：

随机化，首先满足题意的图应该有很多个，所以瞎随……

注意随机的时候，用一种保证每个点的度数最多为2的方式随机就好了

代码

#include
     
      #include
      
       #include
       #include
        
         #include
         
          #include
          
           using namespace std;map
           
            
              ,int>H;int main(){ srand(time(NULL)); int n,m; scanf("%d%d",&n,&m); vector
             
               Q; for(int i=1;i<=n;i++) Q.push_back(i); for(int i=1;i<=m;i++) { int x,y;scanf("%d%d",&x,&y); if(x>y)swap(x,y); H[make_pair(x,y)]=1; } int tot = 0; while(1) { tot++; random_shuffle(Q.begin(),Q.end()); int flag = 0; for(int i=0;i
              
               p) swap(k,p); if(H[make_pair(k,p)]) { flag=1; break; } } if(flag==0) break; if(tot==100) { printf("-1\n"); return 0; } } for(int i=0;i

转载地址：http://nyppa.baihongyu.com/

你可能感兴趣的文章