51nod 1677 treecnt(思维)

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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-
# define MOD
# define INF
# 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<<,l,mid
# define rch p<<|,mid+,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=,f=; char ch=getchar();
while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}
return x*f;
}
inline void Out(int a) {
if(a<) {putchar('-'); a=-a;}
if(a>=) Out(a/);
putchar(a%+'');
}
const int N=;
//Code begin...struct Edge{int p, next;}edge[N<<];
int head[N], cnt=, 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 = , y = ;
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 == ? (x+p)%p : -;
}
void init(){
fac[k]=;
FOR(i,k+,n) fac[i]=fac[i-]*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]=;
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,,n) u=Scan(), v=Scan(), add_edge(u,v), add_edge(v,u);
dfs(,);
printf("%lld\n",(ans+MOD)%MOD);
return ;
}