# Code Forces Bear and Forgotten Tree 3 639B

B. Bear and Forgotten Tree 3 time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output A tree is a connected undirected graph consisting of n vertices and n  -  1 edges. Vertices are numbered 1 through n.

Limak is a little polar bear and Radewoosh is his evil enemy. Limak once had a tree but Radewoosh stolen it. Bear is very sad now because he doesn’t remember much about the tree — he can tell you only three values n, d and h:

The tree had exactly n vertices. The tree had diameter d. In other words, d was the biggest distance between two vertices. Limak also remembers that he once rooted the tree in vertex 1 and after that its height was h. In other words, h was the biggest distance between vertex 1 and some other vertex. The distance between two vertices of the tree is the number of edges on the simple path between them.

Help Limak to restore his tree. Check whether there exists a tree satisfying the given conditions. Find any such tree and print its edges in any order. It’s also possible that Limak made a mistake and there is no suitable tree – in this case print “-1”.

Input The first line contains three integers n, d and h (2 ≤ n ≤ 100 000, 1 ≤ h ≤ d ≤ n - 1) — the number of vertices, diameter, and height after rooting in vertex 1, respectively.

Output If there is no tree matching what Limak remembers, print the only line with “-1” (without the quotes).

Otherwise, describe any tree matching Limak’s description. Print n - 1 lines, each with two space-separated integers – indices of vertices connected by an edge. If there are many valid trees, print any of them. You can print edges in any order.

Examples input 5 3 2 output 1 2 1 3 3 4 3 5 input 8 5 2 output -1 input 8 4 2 output 4 8 5 7 2 3 8 1 2 1 5 6 1 5

```#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 150000
int n,d,h;
int v[MAX+5];
int tag[MAX+5];
int main()
{
scanf("%d%d%d",&n,&d,&h);
if(d>2*h||d<h||(d==1&&h==1&&n!=2))
{printf("-1\n");return 0;}
int i;
memset(v,-1,sizeof(v));
memset(tag,0,sizeof(tag));
for( i=1;i<=h;i++)
{
v[i]=i+1;
tag[i]=1;tag[i+1]=1;
cout<<i<<" "<<v[i]<<endl;
}
if(d-h>0)
{
v=i-1+2;
cout<<1<<" "<<v<<endl;
tag[i-1+2]=1;
}

for(int j=i+2-1;j<d-h-1+i+2-1;j++)
{
v[j]=j+1;
cout<<j<<" "<<v[j]<<endl;
tag[j]=1;tag[j+1]=1;
}
int num;
if(h==1)
num=1;
else
num=2;
for(int j=1;j<=n;j++)
{
if(tag[j]==0)
{
cout<<num<<" "<<j<<endl;
}
}
return 0;

}```

0 条评论

• ### HDU 1103 Flo's Restaurant(模拟+优先队列)

Flo's Restaurant Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/...

• ### PAT 甲级 1078 Hashing

1078. Hashing (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 ...

• ### POJ--1690 (Your)((Term)((Project)))（字符串处理）

(Your)((Term)((Project))) Time Limit: 1000MS Memory Limit: 10000K Total...

• ### C++20 New Features

New keywords: char8_t, co_await, co_return, co_yield, concept, consteval, consti...

• ### ORA-000845 与 /dev/shm(tempfs)

MEMORY_TARGET参数在Oracle 11g被引进，主要是用于控制Oracle对于系统内存的使用，首次将SGA与PGA整合到一起实现自动管理。一...

• ### 用强化学习模拟进化 （CS AI）

进化在地球上产生了人类和动物的智慧。我们认为，发展人工类人智能的道路将通过模拟自然界中的进化过程。在自然界中，驱动大脑发展的过程有两个：进化和学习。进化是缓慢的...

• ### （摘抄）GO语言中template的用法

When a web service responds with data or html pages, there is usually a lot of c...

• ### 4.4 Bond Risk 债券风险

Interest rate factor 是影响利率曲线上各个独立利率的random variables

• ### How to create a CDS redirect view for a given database table

Suppose we have a database table A, and then we create a CDS redirect view B for...

### 活动推荐 