[題目鏈接]

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

[算法]

當d > n - 1或h > n - 1時 ， 無解

當2h < d時無解

當d = 1 , n不為2時 ， 無解

否則 ， 我們先構造一條長度為h的鏈 ， 然後 ， 將一條(d - h)的鏈接到根上 ， 再將剩余節點接到根上

時間復雜度 : O(N)

[代碼]

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

[Codeforces 639B] Bear and Forgotten Tree 3