C. Bear and Forgotten Tree 3
题目连接：
http://www.codeforces.com/contest/658/problem/C
Description
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.
Sample Input
5 3 2
Sample Output
1 2
1 3
3 4
3 5
Hint
题意
你需要构造一棵树，这棵树有n个点，直径为d，深度为h。
题解：
首先构造一个长度为d的链，然后把其中一个距离边上为h的点变为根。
然后我们就不停的在距离根为h上面的那一点不停的加点就好了，使得新加入的点的距离也为h。
为什么呢？想一想10 4 4这个数据就好了。
有一些坑：
4 2 1
10 1 1
把这些数据都过了，应该就没问题了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+7;
int n,d,h;
int ans1[maxn],ans2[maxn];
int tot = 0;
int idx[maxn];
vector<int> T;
int now = 0;
void addans(int x,int y)
{
ans1[tot]=x,ans2[tot]=y;
tot++;
}
int main()
{
scanf("%d%d%d",&n,&d,&h);
if(d>2*h)return puts("-1"),0;
if(d==1&&n>2)return puts("-1"),0;
if(d==2&&h==1)
{
for(int i=2;i<=n;i++)addans(1,i);
for(int i=0;i<tot;i++)printf("%d %d
",ans1[i],ans2[i]);
}
else
{
for(int i=2;i<=n;i++)T.push_back(i);
for(int i=1;i<=h;i++)idx[i]=T[now++];idx[h+1]=1;
for(int i=h+2;i<=d+1;i++)idx[i]=T[now++];
for(int i=1;i<=d;i++)addans(idx[i],idx[i+1]);
for(int i=now;i<T.size();i++)addans(T[i],idx[h]);
for(int i=0;i<tot;i++)printf("%d %d
",ans1[i],ans2[i]);
}
}