Perfect Tree（图论）

题目描述
Given a positive integer k, we define a rooted tree to be k-perfect, if and only if it meets both conditions below:
•Each node is either a leaf node or having exactly k direct offsprings.
•All leaf nodes have the same distance to the root (i.e., all leaf nodes are of the same depth).
Now you are given an unrooted tree, and you should answer these questions:
•Is it possible to assign it a root, so that the tree becomes k-perfect for some positive integer k?
•If possible, what is the minimal k?

输入
Read from the standard input.
Each input contains multiple test cases.
The first line contains a single positive integer T, indicating the number of test cases.
For each test case, its first line contains a positive integer n, describing the number of tree nodes. Each of the next n − 1 lines contains two space-separated integers u and v, which means there exists an edge between node u and v on the tree.
It is guaranteed each test case gives a valid unrooted tree, and the nodes are numbered with consecutive integers from 1 to n.
The sum of n in each input will not exceed 10⁶.

输出
Write to the standard output.
For each test case, output a single integer in a line:
•If the answer to the first question is “No”, output −1.
•Otherwise, output the minimal k.

样例输入
2
7
1 4
4 5
3 1
2 3
7 3
4 6
7
1 4
1 5
1 2
5 3
5 6
5 7

样例输出
2
-1

思路
找树的根节点，从根节点向叶节点推，确认每个节点的子节点是否为k个，叶节点的深度是否相同，注意n=1时答案为1，初始化图时不要用memset。

代码实现

#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>using namespace std;
const int N=1000005;
typedef long long ll;
struct node
{int to,index;
}g[N<<1];
int n;
int head[N<<1],cnt,deg[N];
int dep[N];inline void add(int u,int v)
{g[++cnt].to=v;g[cnt].index=head[u];head[u]=cnt;g[++cnt].to=u;g[cnt].index=head[v];head[v]=cnt;
}void dfs(int u,int nxt,int d)
{dep[u]=d;for(int i=head[u];i;i=g[i].index){int v=g[i].to;if(v==nxt) continue;dfs(v,u,d+1);}
}
int main()
{int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i=0;i<=n;i++) head[i]=deg[i]=0;cnt=0;for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);add(u,v);deg[u]++;deg[v]++;}if(n<=3){puts("1");continue;}int maxl=0;bool flag=false;for(int i=1;i<=n;i++){if(deg[i]==n-1){printf("%d\n",deg[i]);flag=true;break;}maxl=max(maxl,deg[i]);}if(flag) continue;int k=maxl-1,rt;for(int i=1;i<=n;i++){if(deg[i]==k){flag=true;rt=i;break;}}if(!flag){puts("-1");continue;}flag=false;dfs(rt,-1,0);int mosd=-1;for(int i=1;i<=n;i++){if(i!=rt){if(deg[i]==1){if(mosd==-1) mosd=dep[i];else if(mosd!=dep[i]){flag=true;break;}}else{if(deg[i]!=k+1){flag=true;break;}}}else{if(deg[i]!=k){flag=true;break;}}}if(flag) puts("-1");else printf("%d\n",k);}return 0;
}

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