Codeforces Round #236 (Div. 2)C. Searching for Graph

C. Searching for Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 a_i, b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — 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 test(s)

input

16 0

output

1 21 31 41 51 62 32 42 52 63 43 53 6

交题成功了，但是我到现在还是没明白什么意思
题意就是保证子图满足三个条件
我就是解方程搞的，解k*(k-1)/2 <= 2*k+p
找到第一个不满足的k
然后就可以输出答案了，输出的条件就是第k个点不能和k之后的点相连
同时保证输出2*n+p行
感觉最终判题很有可能跪了。。。因为我真的不知道这个题到底是什么意思啊

代码如下：

#include <algorithm>#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <vector>#include <map>#define  MAXN 30#define  INF 0x7FFFFFFF#define  ll long longusing namespace std;int a[MAXN][MAXN];int main(void){int T;cin >> T;while(T--){int n, p;cin >> n >> p;int flag = 0;for(int i=1; i<=n; ++i){int sum = i*i-2*i-2*p;if(sum > 0){flag = i;}}int count = 0;int t = 0;for(int i=1; i<=flag; ++i){for(int j=i+1; j<=n; ++j){cout << i << " " << j << endl;count++;if(count == 2*n+p){t = 1;break;}}if(t == 1) break;}} return 0;}