Codeforces Gym 100269K Kids in a Friendly Class 构造题

Kids in a Friendly Class

题目连接：

http://codeforces.com/gym/100269/attachments

Description

Kevin resembles his class in primary school. There were girls and boys in his class. Some of them were
friends, some were not. But if one person considered another person a friend, the opposite was also true.
Interestingly, every girl had exactly a friends among girls and exactly b friends among boys, whereas
every boy had exactly c friends among girls and exactly d friends among boys.
Kevin does not remember the size of his class. Help him reconstruct the class with minimal possible
number of kids, such that the above conditions are satisfied.

Input

The only line contains four integers a, b, c, and d (1 ≤ a, b, c, d ≤ 50)

Output

Output an example of a class of minimal possible size satisfying the above conditions.
The first line should contains two positive integers: m — the number of girls, and n — the number of
boys.
Let’s assign numbers 1 through m to the girls and m + 1 through m + n to the boys.
Each of the next lines should contain a pair of distinct integers describing a pair of friends by their
numbers. Each pair of friends should appear exactly once in this list.

Sample Input

1 2 1 2

Sample Output

2 4
1 2
1 3
1 5
2 4
2 6
3 4
3 5
4 6
5 6

Hint

题意

每个女生认识a个女生，b个男生

每个男生认识c个女生，d个男生

问你怎么构图，才能使得男生+女生最少

题解：

首先我们假设知道了男生和女生的数量的话

建边就很简单，贪心去建边就好了，每次连接度数最小的点

至于怎么知道男生和女生的数量呢？

首先男生和女生的数量肯定是lcm(b,c)的倍数

然后不断check就好了

check主要只判断同性之间就好了

每条边必须连接两个点之内的check一下就好了

有个定理叫做Havel-Hakimi定理

代码

#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
int a,b,c,d;
bool check(int n,int m)
{
    if(b>m)return false;
    if(c>n)return false;
    if(a>=n)return false;
    if(d>=m)return false;
    if(n%a!=0)return false;
    if(m%b!=0)return false;
    return true;
}
void getedge(int n,int m)
{
    priority_queue<pair<int,int> >Q;
    for(int i=1;i<=n;i++)
        Q.push(make_pair(a,i));
    while(!Q.empty())
    {
        pair<int,int> now = Q.top();
        Q.pop();
        for(int i=0;i<now.first;i++)
        {
            pair<int,int> next = Q.top();
            Q.pop();
            printf("%d %d\n",now.second,next.second);
            next.first--;
            Q.push(next);
        }
    }

    for(int i=n+1;i<=n+m;i++)
        Q.push(make_pair(d,i));
    while(!Q.empty())
    {
        pair<int,int> now = Q.top();
        Q.pop();
        for(int i=0;i<now.first;i++)
        {
            pair<int,int> next = Q.top();
            Q.pop();
            printf("%d %d\n",now.second,next.second);
            next.first--;
            Q.push(next);
        }
    }

    priority_queue<pair<int,int> >Q1;
    priority_queue<pair<int,int> >Q2;
    for(int i=1;i<=n;i++)
        Q1.push(make_pair(b,i));
    for(int i=n+1;i<=m+n;i++)
        Q2.push(make_pair(c,i));
    while(!Q1.empty())
    {
        pair<int,int> now = Q1.top();
        Q1.pop();
        for(int i=0;i<now.first;i++)
        {
            pair<int,int> next = Q2.top();
            Q2.pop();
            printf("%d %d\n",now.second,next.second);
            next.first--;
            if(next.first!=0)Q2.push(next);
        }
    }
}
int main()
{
    freopen("kids.in","r",stdin);
    freopen("kids.out","w",stdout);
    scanf("%d%d%d%d",&a,&b,&c,&d);
    int t=gcd(b,c);
    int x=b/t,y=c/t;
    int n,m;
     for(n=b,m=c;n<=d||m<=a||(m&1)&&(a&1)||(n&1)&&(d&1);n+=x,m+=y);
    swap(n,m);
    cout<<n<<" "<<m<<endl;
    getedge(n,m);
}