Codeforces Round #236 (Div. 2)C. Searching for Graph-优快云博客

本文链接：https://blog.youkuaiyun.com/u011639256/article/details/21442557

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; $\text{[math]}$ ) — 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 
        
1
0

          output 
        
2
3
4
5
6
3
4
5
6
4
5
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 long
using 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;
}