|\ /|
| \ e / |
|L v-----w R|
| / \ |
|/ \|
递归计算L以v为根和R以w为根的分层计数。
包括v的子树由包括v的L中的子树、可选的e和包括w的R中的子树组成。不包括v的子树由不包括v的L中的子树或R中的子树（重复计数空树）组成。这意味着我们可以通过将L的分层计数与R的分层计数进行卷积来获得分层计数。
以下是在您的示例上如何工作的方法。让我们选择根1。
e
1---2---3
1
2---3
这是我对David Eisenstat方案的Python实现：
from sys import stdin
from numpy import *
from scipy import *
def roundup_pow2(x):
"""
Round up to power of 2 (obfuscated and unintentionally faster :).
"""
while x&(x-1):
x = (x|(x>>1))+1
return max(x,1)
def to_long(x):
return long(rint(x))
def poly_mul(a,b):
n = len(a) + len(b) - 1
nr = roundup_pow2(n)
a += [0L]*(nr-len(a))
b += [0L]*(nr-len(b)) # pad with zeros to length n
u = fft(a)
v = fft(b)
w = ifft(u*v)[:n].real # ifft == inverse fft
return map(to_long,w)
def pad(l,s) :
return l+[0L]*(s-len(l))
def make_tree(l,x,y):
l[x][y]=y
l[x].pop(y)
for child in l[x]:
make_tree(l,child,x)
def cut_tree(l,x) :
if len(l[x])==0:
return [1L],[1L]
y,_ = l[x].popitem()
ai,ax=cut_tree(l,x)
bi,bx=cut_tree(l,y)
ci=[0L]+ai
tmp=poly_mul(ai,bi)
padlen=max(len(ci),len(tmp))
ci=pad(ci,padlen)
tmp=pad(tmp,padlen)
ci=map(add,ci,tmp)
cx=[0L]+bi
padlen=max(len(cx),len(bx),len(ax))
cx=pad(cx,padlen)
bx=pad(bx,padlen)
ax=pad(ax,padlen)
tmp=pad([-1],padlen)
cx=map(add,cx,bx)
cx=map(add,cx,ax)
cx=map(add,cx,tmp)
return ci,cx
n,k = map(int,raw_input().split())
l=[{}]
for i in range(1,n+1):
d={}
l.append(d)
for i in range(1,n):
x,y = map(int,raw_input().split())
l[x][y]=y
l[y][x]=x
make_tree(l,1,0)
i,x = cut_tree(l,1)
padlen=max(len(i),len(x))
i=pad(i,padlen)
x=pad(x,padlen)
combined=map(add,i,x)
sum=0L
for i in range(0,k+1) :
sum+=combined[i]
print sum
让我们创建一个稍微大一点的树，就像下面这样。
1
/ | \
2 3 \
/ 4
7 / \
5 6
在编程中，可以通过固定给定子节点的“j”边，然后计算将“k-j”边分配给其他子节点所创建的树的数量来实现。
例如，在上述树中：
F(1, 3) = F(2, 3) + F(3, 3) + F(4, 3) + // we pass k as-is to child
F(2,1)*F(3,1)*F(4,1) + F(2,1)*F(3,2) + F(2,1)*F(4,2) + //consume 1 edge by 2 and distribute 2 to other children
F(2, 2)*F(3,1) + F(2,2)*F(4,1) + // consume 2 edges from node '2' and 1 for other children
F(3,1)*F(4,2)
如上所述，我们将节点2的'r'边固定，然后将'3-r'边分配给其他子节点。 我们对'1'的所有子节点都这样做。
此外，当我们从父节点中分离一个节点时，我们会创建子树。
例如，在上面的情况下，当我们计算F(1, 3)时，我们创建以下
分离的树。
detached_tree += F(2, 2) + F(3, 2) + F(4, 2)
在这里，我们假设一个边被用于从父节点中分离子节点，
并且在子节点中，如果我们使用'k-1'条边，我们将创建F(child, k-1)个子树。
这些树被计数并单独存储在detached_trees中。
一旦我们计算出了所有节点的F(a,k)。
所有子树为'SUM(F(root, k)) for all k' + 'total nodes - 1' + detached_trees。
我们将'total nodes - 1'添加到总数中。这是因为当一个节点（除了根节点）被分离 从树中，它会创建两个缺少1条边的树。虽然其中一个树被计算 在F(parent, 1)中，另一个树没有被计算在任何地方，因此需要在总数中计算。
以下是上述算法的C代码。递归可以进一步优化。
#define MAX 51
/* We use the last entry of alist to store number of children of a given node */
#define NUM_CHILD(alist, node) (alist[node][MAX])
int alist[MAX][MAX+1] = {0};
long F[MAX][MAX]={0};
long detached_subtrees = 0;
/*
* We fix one of the child node for 'i' edges out of 'n', then we traverse
* over the rest of the children to get 'n-i' edges, we do so recursivly.
* Note that if 'n' is 1, we can always build a subtree by detaching.
*/
long REST_OF_NODES_SUM(int node, int q, int n)
{
long sum = 0, i, node2, ret = 0, nd;
/* fix node2 and calcualte the subtree for rest of the children */
for(nd = q; nd < NUM_CHILD(alist, node); nd++) {
node2 = alist[node][nd];
/* Consume 'i' edges and send 'n-i' for other children of node */
for (i = 1; i < n ; i++) {
sum = REST_OF_NODES_SUM(node, nd + 1, n - i);
ret += (F[node2][i] * sum);
/* Add one for 'node2' getting detached from tree */
if (i == 1) { ret += sum; }
}
ret += F[node2][n];
/* If only one edge is to be consumed, we detach 'node2' from the tree */
if (n == 1) { ret++; }
}
return ret;
}
void get_counts(int N, int K, int node, int root)
{
int child_node;
int i, j, p, k;
if (NUM_CHILD(alist, node) == 0) { return; }
for(i = 0 ; i < NUM_CHILD(alist, node); i++) {
child_node = alist[node][i];
/* Do a recursive traversal of all children */
get_counts(N, K, child_node, node);
F[node][1] += (F[child_node][1]);
}
F[node][1] += NUM_CHILD(alist, node);
for (k = 2; k <= K; k++) {
for(p = 0; p < NUM_CHILD(alist, node); p++) {
child_node = alist[node][p];
F[node][k] += F[child_node][k];
/* If we remove this child, then we create subtrees in the child */
detached_subtrees += F[child_node][k-1];
/* Assume that 'child_node' is detached, find tree created by rest
* of children for 'k-j' edges */
F[node][k] += REST_OF_NODES_SUM(node, p + 1, k - 1);
/* Fix one child node for 'j' edges out of 'k' and traverse over the rest of
* children for 'k - j' edges */
for (j = 1; j < k ; j++) {
if (F[child_node][j]) F[node][k] += (F[child_node][j] * REST_OF_NODES_SUM(node, p + 1, k - j));
}
}
}
}
void remove_back_ref(int parent, int node)
{
int c;
for (c = 0; c < NUM_CHILD(alist, node); c++) {
if (alist[node][c] == parent) {
if ((c + 1) == NUM_CHILD(alist, node)) {
NUM_CHILD(alist, node)--;
alist[node][c] = 0;
} else {
/* move last entry here */
alist[node][c] = alist[node][NUM_CHILD(alist, node)-1];
alist[node][NUM_CHILD(alist, node)-1] = 0;
NUM_CHILD(alist, node)--;
}
}
}
}
/* go to each child and remove back links */
void normalize(int node)
{
int j, child;
for (j = 0; j < NUM_CHILD(alist, node); j++) {
child = alist[node][j];
remove_back_ref(node, child);
normalize(child);
}
}
long cutTree(int N, int K, int edges_rows, int edges_columns, int** edges)
{
int i, j;
int node, index;
long ret = 0;
/* build an adjacency list from the above edges */
for (i = 0; i < edges_rows; i++) {
alist[edges[i][0]][NUM_CHILD(alist, edges[i][0])] = edges[i][1];
alist[edges[i][1]][NUM_CHILD(alist, edges[i][1])] = edges[i][0];
NUM_CHILD(alist, edges[i][0])++;
NUM_CHILD(alist, edges[i][1])++;
}
/* get rid of the back links in children */
normalize(1);
get_counts(N, K, 1, 1);
for (i = 1; i <= K; i++) { ret += F[1][i]; }
/* Every node (except root) when detached from tree, will create one extra subtree. */
ret += (N - 1);
/* The subtrees created by detaching from parent */
ret += detached_subtrees;
/* Add two for empty and full tree */
ret += 2;
return ret;
}
main(int argc, char *argv[])
{
int **arr;
int ret, i, N, K, x, y;
scanf("%d%d", &N, &K);
arr = malloc((N - 1) * sizeof(int*));
for (i = 0; i < (N - 1); i++) { arr[i] = malloc(2*sizeof(int)); }
for (i = 0; i < N-1; i++) { scanf("%d%d", &x, &y); arr[i][0] = x; arr[i][1] = y; }
printf("MAX %d ret %ld\n", MAX, cutTree(N, K, N-1, 2, arr));
}
{2}？补语也必须是一棵树吗？ - ypercubeᵀᴹ