codeforces 658C C. Bear and Forgotten Tree 3(tree+乱搞)

题目链接：C. Bear and Forgotten Tree 3time limit per test
2 secondsmemory limit per test
256 megabytesinput
standard inputoutput
standard outputA 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.Examplesinput

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

题意:

给一棵树的节点数目n,直径d,最大深度h,问是否存在一棵树满足这些条件,存在的话任意输出一棵树;
比赛时先是被人hack,发现了一处错误改了后又交了一发,但最后还是挂在了system test上;

思路：

两种情况不存在树,一种是d>2*h,一种是d==1&&h==1&&n>2时;可以先生成最大深度的一条链,再在次基础上把最大直径的链生成;其余的都和最大深度的节点的父节点相连接就满足要求了;

AC代码：

/*
2014300227    658C - 38    GNU C++11    Accepted    46 ms    2176 KB
*/

#include <bits/stdc++.h>
using namespace std;
int n,d,h;
int main()
{
scanf("%d%d%d",&n,&d,&h);
if(d>2*h||(h==1&&d==1&&n>2))cout<<"-1"<<"\n";
else
{
for(int i=1;i<=h;i++)
{
printf("%d %d\n",i,i+1);
}
if(d==h)
{
if(d<n-1)
{
for(int i=h+2;i<=n;i++)
{
printf("%d %d\n",h,i);
}
}
}
else
{

cout<<"1"<<" "<<h+2<<"\n";
for(int i=h+2;i<=d;i++)
{
printf("%d %d\n",i,i+1);
}
for(int i=d+2;i<=n;i++)
{
printf("%d %d\n",h,i);
}

}

}

return 0;
}