F. K-Set Tree time limit per test3 seconds memory limit per test512 megabytes inputstandard input outputstandard output You are given a tree G with n vertices and an integer k. The vertices of the tree are numbered from 1 to n. For a vertex r and a subset S of vertices of G, such that |S|=k, we define f(r,S) as the size of the smallest rooted subtree containing all vertices in S when the tree is rooted at r. A set of vertices T is called a rooted subtree, if all the vertices in T are connected, and for each vertex in T, all its descendants belong to T. You need to calculate the sum of f(r,S) over all possible distinct combinations of vertices r and subsets S, where |S|=k. Formally, compute the following: ∑r∈V∑S⊆V,|S|=kf(r,S), where V is the set of vertices in G. Output the answer modulo 109+7. Input The first line contains two integers n and k (3≤n≤2⋅105, 1≤k≤n). Each of the following n−1 lines contains two integers x and y (1≤x,y≤n), denoting an edge between vertex x and y. It is guaranteed that the given edges form a tree. Output Print the answer modulo 109+7. Examples inputCopy 3 2 1 2 1 3 outputCopy 25 inputCopy 7 2 1 2 2 3 2 4 1 5 4 6 4 7 outputCopy 849 Note The tree in the second example is given below: We have 21 subsets of size 2 in the given tree. Hence, S∈{{1,2},{1,3},{1,4},{1,5},{1,6},{1,7},{2,3},{2,4},{2,5},{2,6},{2,7},{3,4},{3,5},{3,6},{3,7},{4,5},{4,6},{4,7},{5,6},{5,7},{6,7}}. And since we have 7 vertices, 1≤r≤7. We need to find the sum of f(r,S) over all possible pairs of r and S. Below we have listed the value of f(r,S) for some combinations of r and S. r=1, S={3,7}. The value of f(r,S) is 5 and the corresponding subtree is {2,3,4,6,7}. r=1, S={5,4}. The value of f(r,S) is 7 and the corresponding subtree is {1,2,3,4,5,6,7}. r=1, S={4,6}. The value of f(r,S) is 3 and the corresponding subtree is {4,6,7}.
思路：
- 这是一道算节点贡献的题
- 贡献可以分为2中情况 为 根，和不为根， 但是这个点代表的2种贡献，都是以这个点为最近的公共祖先（最小的子节点）
- 2种情况 都是 分别从他的儿子树们，选一些节点出来， 不能所有点都选一个儿子的，这样就不能保证最近公共祖先
- 那么怎么办内？，就利用减法思维 C（sz，K） - C（各个儿子的size，K，）， 这样就一定能保证上面的条件，
- 根 就 上面的 xN；
- 不是根，就枚举 相邻的节点，把他根， 然后 种类数X剩下的szX父亲的数量
- 种类数： 同理可得 C（sz，K） - C（各个儿子的size，K，），当然这里用 减法 优化时间复杂度，不然就n^2了；
- 先把 C（各个相邻节点的size，K，） 求和， 然后 这个和 - C（父亲size，K）
- 具体看代码（很简单）
核心：就是 组合数减法，可以让最小公共祖先成立
#include <bits/stdc++.h> using namespace std; #define ri register int #define M 200005 template <class G> void read(G &x) { x=0;int f=0;char ch=getchar(); while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();} x=f?-x:x; return ; } long long ans=0; const int mod=1e9+7; long long n,m; vector<int> p[M]; int vis[M]; int sz[M]; long long inv[M],arr[M]; long long qsn(long long a,int n) { long long ans=1; while(n) { if(n&1) ans=ans*a%mod; n>>=1;a=a*a%mod; } return ans; } long long zh(long long a,long long b) { if(a==b||b==0) return 1; if(a<b||a==0) return 0; return arr[a]*inv[a-b]%mod*inv[b]%mod; } void init(){ arr[0]=1;inv[0]=1; for(ri i=1;i<=n;i++) { arr[i]=i*arr[i-1]%mod; inv[i]=qsn(arr[i],mod-2); } } void dfs1(int a) { vis[a]=1; sz[a]++; for(ri i=0;i<p[a].size();i++) { int b=p[a][i]; if(vis[b]) continue; dfs1(b); sz[a]+=sz[b]; } return ; } void dfs2(int a) { vis[a]=1; ans=(ans+zh(n,m)*n%mod)%mod; // geng long long tmp=0; for(ri i=0;i<p[a].size();i++) { int b=p[a][i]; if(vis[b]==0) { ans=(ans-zh(sz[b],m)*n%mod+mod)%mod; tmp=(tmp+zh(sz[b],m))%mod; } else { ans=(ans-zh(n-sz[a],m)*n%mod+mod)%mod; tmp=(tmp+zh(n-sz[a],m))%mod; } } for(ri i=0;i<p[a].size();i++) { int b=p[a][i]; if(vis[b]==0) { ans=(ans+zh(n-sz[b],m)*(n-sz[b])%mod*sz[b])%mod; ans=(ans-(tmp-zh(sz[b],m))*(n-sz[b])%mod*sz[b]%mod+mod)%mod; } else { ans=(ans+zh(sz[a],m)*(sz[a])%mod*(n-sz[a]))%mod; ans=(ans-(tmp-zh(n-sz[a],m))*(sz[a])%mod*(n-sz[a])%mod+mod)%mod; } } for(ri i=0;i<p[a].size();i++) { int b=p[a][i]; if(vis[b]) continue; dfs2(b); } return ; } int main(){ read(n); read(m); for(ri i=1;i<n;i++) { int a,b; read(a);read(b); p[a].push_back(b); p[b].push_back(a); } init(); dfs1(1); memset(vis,0,sizeof(vis)); dfs2(1); printf("%lld",ans); return 0; }