codeforces Searching for Graph

来源:互联网 发布:python做网站的服务器 编辑:程序博客网 时间:2024/05/22 00:52

Description

Let’s call an undirected graph of n vertices p-interesting, if the following conditions fulfill:
• the graph contains exactly 2n + p edges;
• the graph doesn’t contain self-loops and multiple edges;
• for any integer k (1 ≤ k ≤ n), any subgraph consisting of k vertices contains at most 2k + p edges.

A subgraph of a graph is some set of the graph vertices and some set of the graph edges. At that, the set of edges must meet the condition: both ends of each edge from the set must belong to the chosen set of vertices.

Your task is to find a p-interesting graph consisting of n vertices.

Input

The first line contains a single integer t (1 ≤ t ≤ 5) — the number of tests in the input. Next t lines each contains two space-separated integers: n, p (5 ≤ n ≤ 24; p ≥ 0; ) — the number of vertices in the graph and the interest value for the appropriate test.

It is guaranteed that the required graph exists.

Output

For each of the t tests print 2n + p lines containing the description of the edges of a p-interesting graph: the i-th line must contain two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — two vertices, connected by an edge in the resulting graph. Consider the graph vertices numbered with integers from 1 to n.

Print the answers to the tests in the order the tests occur in the input. If there are multiple solutions, you can print any of them.

Sample Input

Input
1
6 0

Output
1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6

题意:给定n 和p ,我们需要构造一张点数为n ,边数为2n+p 的简单无向图,满足任意一个点数为k 的子图的边数不超过2k+p 。
建议参考:http://blog.csdn.net/popoqqq/article/details/45749329

#include <iostream>#include <cmath>using namespace std;int main(){    int t;    cin>>t;    for(int k=0;k<t;k++)    {        int n,p,num=0;        cin>>n>>p;        int sum=2*n+p;        for(int i=1;i<=n&&num<sum;i++)        {            for(int j=i+1;j<=n&&num<sum;j++)            {                num++;                cout<<i<<" "<<j<<endl;            }        }    }    return 0;}
0 0
原创粉丝点击