[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 = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}

int main()
{
        
        int n , h , d;
        read(n); read(d); read(h);
        if (d > n - 1 || h > n - 1)
        {
                printf("-1
");
                return 0;
        }
        if (h * 2 < d) 
        {
                printf("-1
");
                return 0;        
        }
        if (d == 1 && n != 2)
        {
                printf("-1
");
                return 0;
        }
        for (int i = 2; i <= h + 1; i++) printf("%d %d
",i,i - 1);
        for (int i = 1; i <= d - h; i++) 
        {
                if (i == 1) printf("%d %d
",1,h + 1 + i);
                else printf("%d %d
",h + i,h + 1 + i);
        }
        if (d == h)
        {
                for (int i = d + 2; i <= n; i++) 
                        printf("%d %d
",2,i);
        } else
        {
                for (int i = d + 2; i <= n; i++)
                        printf("%d %d
",1,i);
        }
        
        return 0;
    
}