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codeforces Searching for Graph

2015-11-06 20:39 369 查看
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:/article/2348635.html

[code]#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;
}
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