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[1]=i-1+2;
     cout<<1<<" "<<v[1]<<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;


}

转载于:https://www.cnblogs.com/dacc123/p/8228715.html