codeforces22c System Administrator【给定一个割顶输出边 BCC】

Description

Bob got a job as a system administrator in X corporation. His first task was to connectn servers with the help ofm two-way direct connection so that it becomes possible to transmit
data from one server to any other server via these connections. Each direct connection has to link two different servers, each pair of servers should have at most one direct connection. Y corporation, a business rival of X corporation, made Bob an offer that
he couldn‘t refuse: Bob was asked to connect the servers in such a way, that when server with indexv fails, the transmission of data between some other two servers becomes impossible, i.e. the system stops being connected.
Help Bob connect the servers.

Input

The first input line contains 3 space-separated integer numbers
n, m,
v (3?≤?n?≤?10⁵,?0?≤?m?≤?10⁵,?1?≤?v?≤?n),n — amount of servers,m
— amount of direct connections, v — index of the server that fails and leads to the failure of the whole system.

Output

If it is impossible to connect the servers in the required way, output
-1. Otherwise output m lines with 2 numbers each — description of all the direct connections in the system. Each direct connection is described by two numbers — indexes of two servers, linked by this direct connection.
The servers are numbered from 1. If the answer is not unique, output any.

Sample Input

Input

5 6 3

Output

1 2
2 3
3 4
4 5
1 3
3 5

Input

6 100 1

Output

-1

这又是一个乱搞的题QAQ

题意：已知点数、边数、割点的序号（一直以为是割点的个数，没法求了啊==），求是否能构成满足条件的图，输出

做法：首先要判断是否满足条件，我们需要找一个边数和点数的关系，点数的下限一定是等于边数的，点数的上限可以这么想：既然有一个点是割顶，那么把剩下的点分成两团，分别都是完全图，然后割顶是中间的“纽带”，那么想来这两团的个数相等的时候边数最多，上限可求。我们顺着这个思路就可以按顺序写出边啦~~

#include <iostream>
#include<cstdio>
using namespace std;
int n,m,v;
int main()
{
    scanf("%d%d%d",&n,&m,&v);
    {
        if(m<n-1||m>(n*n-3*n+4)/2)
        printf("-1\n");
        else
        {
            int la;
            for(int i=1;i<=n;i++)
            {
                if(i!=v)
                {
                    printf("%d %d\n",i,v);
                    la=i;
                   // break;
                }
            }
            m-=(n-1);
            for(int i=1;i<la&&m;i++)
            {
                for(int j=i+1;j<la&&m;j++)
                {
                    if(i!=v&&j!=v)
                    {
                        printf("%d %d\n",i,j);
                        m--;
                    }

                }
            }
        }
    }
    return 0;
}

写了一天vtk感觉ACM的代码真是短的感人啊==