Codeforce - 402C - Searching for Graph

C. Searching for Graph
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
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：有 n 个点，要求连线成图，且该图中的边数不超过 2*n+p 条。

AC代码：

#include <iostream>
using namespace std;

int main()
{
    int t;cin>>t;
    while (t--)
    {
        int n,p;cin>>n>>p;

        bool flag=false ;int cnt=0;
        for (int i=1;i<=n;i++)
        {
            for (int j=i+1;j<=n;j++)
            {
                cout<<i<<" "<<j<<endl;
                cnt++;
                if (cnt==2*n+p) {flag=true;break;}
            }
            if (flag) break;
        }
    }
}

/*
其实这条题的难点就在于读题。
*/