[Codeforces 639B] Bear and Forgotten Tree 3

[题目链接]

https://codeforces.com/problemset/problem/639/B

[算法]

当d > n - 1或h > n - 1时 ， 无解

当2h < d时无解

当d = 1 , n不为2时 ， 无解

否则 ， 我们先构造一条长度为h的链 ， 然后 ， 将一条(d - h)的链接到根上 ， 再将剩余节点接到根上

时间复杂度 : O(N)

[代码]

#include<bits/stdc++.h>
using namespace std;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}

int main()
{

        int n , h , d;
        read(n); read(d); read(h);
        if (d > n -  || h > n - )
        {
                printf("-1\n");
                return ;
        }
        if (h *  < d)
        {
                printf("-1\n");
                return ;
        }
        if (d ==  && n != )
        {
                printf("-1\n");
                return ;
        }
        for (int i = ; i <= h + ; i++) printf("%d %d\n",i,i - );
        for (int i = ; i <= d - h; i++)
        {
                if (i == ) printf("%d %d\n",,h +  + i);
                else printf("%d %d\n",h + i,h +  + i);
        }
        if (d == h)
        {
                for (int i = d + ; i <= n; i++)
                        printf("%d %d\n",,i);
        } else
        {
                for (int i = d + ; i <= n; i++)
                        printf("%d %d\n",,i);
        }

        return ;

}