51nod 1677 treecnt(思维)

题意:

给定一棵n个节点的树，从1到n标号。选择k个点，你需要选择一些边使得这k个点通过选择的边联通，目标是使得选择的边数最少。

现需要计算对于所有选择k个点的情况最小选择边数的总和为多少。

 

考虑每条边对答案的贡献，令x为这条边左边的点数，则n-x为这条边右边的点数。

满足条件的情况数=总情况数-不满足条件的情况数。即C(n,k)-(C(x,k)+C(n-x,k)).

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <bitset>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-8
# define MOD 1000000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
inline int Scan() {
    int x=0,f=1; char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
inline void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=100005;
//Code begin...

struct Edge{int p, next;}edge[N<<1];
int head[N],  cnt=1, siz[N];
LL fac[N], ans;
int n, k;

void exgcd(LL a,LL b,LL & d,LL & x,LL & y){
    if(!b) d = a, x = 1, y = 0;
    else exgcd(b, a%b, d, y, x), y -= x*(a/b);
}
LL inv(LL a, LL p){
    LL d, x, y;
    exgcd(a, p, d, x, y);
    return d == 1 ? (x+p)%p : -1;
}
void init(){
    fac[k]=1;
    FOR(i,k+1,n) fac[i]=fac[i-1]*i%MOD*inv(i-k,MOD)%MOD;
}
void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}
void dfs(int x, int fa){
    int tmpx, tmpy;
    siz[x]=1;
    for (int i=head[x]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (v==fa) continue;
        dfs(v,x); siz[x]+=siz[v];
        tmpx=siz[v]; tmpy=n-siz[v];
        ans=(ans+fac[n]-fac[tmpx]-fac[tmpy])%MOD;
    }
}
int main ()
{
    int u, v;
    n=Scan(); k=Scan();
    init();
    FO(i,1,n)  u=Scan(), v=Scan(), add_edge(u,v), add_edge(v,u);
    dfs(1,0);
    printf("%lld\n",(ans+MOD)%MOD);
    return 0;
}