Equidistant Vertices-树型dp

Posted 2021-08-27 PushyTao

Description
A tree is an undirected connected graph without cycles.

You are given a tree of n vertices. Find the number of ways to choose exactly k vertices in this tree (i. e. a k-element subset of vertices) so that all pairwise distances between the selected vertices are equal (in other words, there exists an integer c such that for all u,v (u≠v, u,v are in selected vertices) du,v=c, where du,v is the distance from u to v).

Since the answer may be very large, you need to output it modulo 109+7.

Input
The first line contains one integer t (1≤t≤10) — the number of test cases. Then t test cases follow.

Each test case is preceded by an empty line.

Each test case consists of several lines. The first line of the test case contains two integers n and k (2≤k≤n≤100) — the number of vertices in the tree and the number of vertices to be selected, respectively. Then n−1 lines follow, each of them contains two integers u and v (1≤u,v≤n, u≠v) which describe a pair of vertices connected by an edge. It is guaranteed that the given graph is a tree and has no loops or multiple edges.

Output
For each test case output in a separate line a single integer — the number of ways to select exactly k vertices so that for all pairs of selected vertices the distances between the vertices in the pairs are equal, modulo 109+7 (in other words, print the remainder when divided by 1000000007).

Example
input

3

4 2
1 2
2 3
2 4

3 3
1 2
2 3

5 3
1 2
2 3
2 4
4 5

output

6
0
1

typedef int itn;
struct node{
	int v,nex;
}e[107 << 1];
int head[107 << 1],cnt;
int n,k;
void init(){
	cnt = 0;
	for(int i=0;i<(107<<1);i++) head[i] = -1;
}
void add(int u,int v){
	e[cnt].v = v;
	e[cnt].nex = head[u];
	head[u] = cnt ++;
}
int siz[107];
bool vis[107];
int dp[107][107];
void dfs(int u,int fa,int dep,int pnt){
	vis[u] = 1;
	if(dep == pnt){
		siz[u] ++;
		return;
	}
	for(int i=head[u];~i;i=e[i].nex){
		int to = e[i].v;
		if(to == fa) continue;
		dfs(to,u,dep+1,pnt);
		siz[u] += siz[to];
	}
}
int main()
{
	int _ = read;
	while(_ --){
		n = read,k = read;
		init();
		for(int i=1;i<n;i++){
			int u = read,v = read;
			add(u,v);
			add(v,u);
		}
		ll ans = 0;
		if(k == 2){
			ans = n * (n - 1) / 2 % mod;
			printf("%lld\\n",ans);
			continue;
		}
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				memset(vis,0,sizeof vis);
				memset(siz,0,sizeof siz);
				memset(dp,0,sizeof dp);
				dfs(i,0,0,j);
				dp[0][0] = 1;
				int it = 0;
				for(int tj = head[i];~tj;tj=e[tj].nex){
					int to = e[tj].v;
					if(siz[to] == 0) continue;
					for(int kk=0;kk<=it;kk++){
						dp[it+1][kk] += dp[it][kk];
						dp[it+1][kk] %= mod;
						dp[it+1][kk+1] += dp[it][kk] * siz[to] % mod;
						dp[it+1][kk+1] %= mod;
					}
					it ++;
				}
				ans += dp[it][k];
				ans %= mod;
			}
		}
		printf("%lld\\n",ans);
	}
    return 0;
}