CodeForces Round#22 C System Administrator 构造割点图

题目描述：

Description
Bob got a job as a system administrator in X corporation. His first task was to connect n servers with the help of m 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 index v 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

题目分析：

题意就是n个点，m条边，其中点v是割点。
构造这个无向图，若不能构造就输出-1。
首先，能否构造这个图，取决于n与m的大小关系。m的最小值就是树的条件m=n-1。最大值就是v点分割之后的两个子图x和y都是完全图。通过推导得到m<=(n*n-3*n+4)/2，即m的最大值。
然后先输出与割点相连的边，在输出其它m-（n-1）条边，但是在输出的时候要保证v点是图的割点。因此不能任意输出。

代码如下：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;

int n,m,v;
int main()
{
    while(scanf("%d%d%d",&n,&m,&v)!=-1)
    {
        if (m<n-1 || m>((n*n-3*n+4)/2))
        {
            printf("-1\n");
            continue;
        }
        for(int i=1; i<=n; i++)
        {
            if (i!=v) printf("%d %d\n",i,v);//n-1条与割点相连的边
        }
        m-=n-1;
        int tt=1;//通过tt保证v是割点
        if (v==1) tt=2;
        for(int i=1; i<=n && m>0; i++)
            if(i!=v && i!=tt)
                for(int j=i+1; j<=n &&m>0; j++)
                    if(j!=v && j!=tt)
                    {
                        printf("%d %d\n",i,j);//其它的边
                        m--;
                    }
    }
    return 0;
}