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: np (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 ≤ nai ≠ 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 test(s)
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

这个题是A题的水平额。。是在考题意么。。

 #include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
const int N=;
using namespace std; int main()
{
int t,n,p;
cin>>t;
while(t--)
{
cin>>n>>p;
int m = *n+p;
int cnt = ,flag = ;
for (int i = ;i <= n; i++)
{
for (int j = i+;j <= n; j++)
{
cout<<i<<" "<<j<<endl;
cnt++;
if (cnt==m)
{
flag = ;
break;
}
}
if (flag)
break;
}
}
return ;
}

C. Searching for Graph(cf)的更多相关文章

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

    题目大意 我们说一个无向图是 p-interesting 当且仅当这个无向图满足如下条件: 1. 该图恰有 2 * n + p 条边 2. 该图没有自环和重边 3. 该图的任意一个包含 k 个节点的子 ...

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

    题目地址 /* 题意:要你构造一个有2n+p条边的图,使得,每一个含k个结点子图中,最多有2*k+p条边 水得可以啊,每个点向另外的点连通,只要不和自己连,不重边就可以,正好2*n+p就结束:) */ ...

  3. SPOJ #442 Searching the Graph

    Just CS rookie practice on DFS\BFS. But details should be taken care of: 1. Ruby implementation got ...

  4. CF_402C Searching for Graph 乱搞题

    题目链接:http://codeforces.com/problemset/problem/402/C /**算法分析: 乱搞题,不明白题目想考什么 */ #include<bits/stdc+ ...

  5. Codeforces Round #236 (Div. 2)

    A. Nuts time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:st ...

  6. Reading task(Introduction to Algorithms. 2nd)

    Introduction to Algorithms 2nd ed. Cambridge, MA: MIT Press, 2001. ISBN: 9780262032933. Introduction ...

  7. CF1082解题报告

    CF1082A Vasya and Book 模拟一下即可 \(Code\ Below:\) #include <bits/stdc++.h> using namespace std; c ...

  8. cf 542E - Playing on Graph

    cf 542E - Playing on Graph 题目大意 给定一个\(n\le 1000\)个点的图 求经过一系列收缩操作后能否得到一条链,以及能得到的最长链是多长 收缩操作: 选择两个不直接相 ...

  9. CF 329C(Graph Reconstruction-随机化求解-random_shuffle(a+1,a+1+n))

    C. Graph Reconstruction time limit per test 3 seconds memory limit per test 256 megabytes input stan ...

随机推荐

  1. TWaver 3D作品Viewer查看器

    为了让开发者更方便的对各类3D模型.设备.物体进行浏览和查看,我们直接封装了mono.Viewer组件.它可以直接根据给定的数据源(json.obj.url等)进行数据加载和浏览展示.对于一般的3D设 ...

  2. 将node-webkit打包后文件用nsis再打包成安装包

  3. 洛谷 1071 潜伏者(NOIp2009提高组)

    [题意概述] 给出三行字符串,前两行代表密码与明文的对应关系,第三行为待翻译的文本.要求按照对应关系翻译文本. [题解] 直接模拟即可. 注意判断Failed的情况. #include<cstd ...

  4. 将网络图片转换为base64

    public static function htmlPdf() { $img_path = Env::get('ROOT_PATH').'/public/images/wechat/user.jpg ...

  5. 关于Scrum 实战故事录播的感悟升级

    昨晚与几位自组织的伙伴进行了<Scrum 实战> 第17 章 <富有成效的每日站会>录播Sprint 不断的优化和精进的感悟. 首先,D兄给予了如下的建议: 1. 将段落 分得 ...

  6. Promise 异步编程

    //1.解决异步回调问题 //1.1 如何同步异步请求 //如果几个异步操作之间并没有前后顺序之分,但需要等多个异步操作都完成后才能执行后续的任务,无法实现并行节约时间 const fs = requ ...

  7. codevs——1436 孪生素数 2

    1436 孪生素数 2  时间限制: 2 s  空间限制: 1000 KB  题目等级 : 白银 Silver 题解       题目描述 Description 如m=100,n=6 则将输出100 ...

  8. nginx: 添加文件下载目录

    修改nginx.conf,添加如下行: location /file/ {    alias /usr/share/nginx/html/file/;    add_header Content-di ...

  9. doT js模板入门 3

    for 循环前推断循环的list是否为空 <script id="invoiceListDot" type="text/x-dot-template"&g ...

  10. 1.7-BGP①

    IGP:   包括RIP/EIGRP/OSPF/ISIS/ODR等动态路由协议   运行在同一个AS中,   通过Cost/Metirc来判断路由的优劣(越小越好):   AS:自治系统(小)   A ...