Codeforces 550D. Regular Bridge 构造

求一个图,每个点的度数都为K而且必须至少要有一个桥.

构造题:

只有k为奇数的时候有解, 构造这样的一个图,左边一团有 k+1 个点 , 右边一团也有 k+1 个点, 中间经过 m1 , m2 连着一个桥.

如果左右两团是完全图,则每个点的度数都为k, 现在考虑如何通过m1,m2连接起来而又不改变度数.

显然这个图是对称的,只考虑左边和点m1,m1和m2是一个桥,要连一条边,m1 和左边的团某个点A要连在一起,又要连一条边,这时A点的度数多了,要和团里的其他点B断掉一条边,为了保持B的度数不变,B再连一条边到m1,这时m1的度就至少为3了,然后将m1的度补全. 每次从左边团中删掉一条边,然后将这条边的两个点连到m1上就可以了.m1的度每次增加2,所以如果k为奇数的话都可以用这样的方法构造出来.

D. Regular Bridge

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

An undirected graph is called k-regular, if the degrees of all its vertices are equal k. An edge of a connected graph is called a bridge, if after removing it the graph is being split into two connected components.

Build a connected undirected k-regular graph containing at least one bridge, or else state that such graph doesn't exist.

Input

The single line of the input contains integer k (1 ≤ k ≤ 100) — the required degree of the vertices of the regular graph.

Output

Print "NO" (without quotes), if such graph doesn't exist.

Otherwise, print "YES" in the first line and the description of any suitable graph in the next lines.

The description of the made graph must start with numbers n and m — the number of vertices and edges respectively.

Each of the next m lines must contain two integers, a and b (1 ≤ a, b ≤ n, a ≠ b), that mean that there is an edge connecting the vertices aand b. A graph shouldn't contain multiple edges and edges that lead from a vertex to itself. A graph must be connected, the degrees of all vertices of the graph must be equal k. At least one edge of the graph must be a bridge. You can print the edges of the graph in any order. You can print the ends of each edge in any order.

The constructed graph must contain at most 106 vertices and 106 edges (it is guaranteed that if at least one graph that meets the requirements exists, then there also exists the graph with at most 106 vertices and at most 106 edges).

Sample test(s)

input

1

output

YES2 11 2

Note

In the sample from the statement there is a suitable graph consisting of two vertices, connected by a single edge.

import java.util.*;import java.math.*;public class Main{int k,midl,midr;boolean[][] edge = new boolean[330][330];boolean[] used = new boolean[330];void cut_and_link(int from,int to,int goal){int tk=k-3;while(tk>0){boolean flag=false;for(int i=from;i<to&&flag==false;i++){if(used[i]==true) continue;for(int j=i+1;j<to;j++){if(i==2*k+4||i==4+k||i==1+k||i==1) continue;if(j==2*k+4||j==4+k||j==1+k||j==1) continue;if(used[j]==true) continue;if(edge[i][j]==true){used[i]=used[j]=true;//cutedge[i][j]=edge[j][i]=false;//linkedge[i][goal]=edge[goal][i]=true;edge[j][goal]=edge[goal][j]=true;// flagflag=true;break;}}}tk-=2;}}void build_tuan(int from ,int to){for(int i=from;i<=to;i++){for(int j=i+1;j<=to;j++){edge[i][j]=edge[j][i]=true;}}}void sovle(int k){midl=k+2; midr=k+3;build_tuan(1,k+1);edge[1][midl]=edge[midl][1]=true;edge[1][k+1]=edge[k+1][1]=false;edge[k+1][midl]=edge[midl][k+1]=true;cut_and_link(2,k+1,midl);edge[midl][midr]=edge[midr][midl]=true;build_tuan(k+4,2*k+4);edge[2*k+4][midr]=edge[midr][2*k+4]=true;edge[k+4][2*k+4]=edge[2*k+4][k+4]=false;edge[k+4][midr]=edge[midr][k+4]=true;cut_and_link(k+4,2*k+4,midr);}Main(){Scanner in = new Scanner(System.in);k=in.nextInt();if(k%2==0){System.out.println("No");return ;}System.out.println("Yes");if(k==1){System.out.println("2 1\n1 2");}else{sovle(k);int num=2*k+4,cnt=0;for(int i=1;i<=num;i++)for(int j=i+1;j<=num;j++)if(edge[i][j]) cnt++;System.out.printf("%d %d\n",num,cnt);for(int i=1;i<=num;i++){for(int j=i+1;j<=num;j++){if(edge[i][j]){System.out.printf("%d %d\n",i,j);}}}}}public static void main(String[] args){new Main();}}