BNUOJ 52305 Around the World 树形dp

题目链接：

https://www.bnuoj.com/v3/problem_show.php?pid=52305

Around the World

Time Limit: 20000ms
Memory Limit: 1048576KB

题意

给你一颗树，相邻两点间有2c条边，问以1为起点和终点的欧拉回路有多少种。

题解

树形dp
兄弟之间考虑可重集的排序，父子之间则考虑下插板法。
dp[u]表示以u为根的子树能跑的所有欧拉回路。

include

include

include

include

include

include

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef unsigned long long ULL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e5+10;
const int maxm=2e6+100;
const int mod=1e9+7;

VPII G[maxn];

int n;

LL dp[maxn];
int cntv[maxn];
LL inv[maxm],invfac[maxm],fac[maxm];

LL get_C(int n,int m){
return fac[n]invfac[m]%modinvfac[n-m]%mod;
}

void dfs(int u,int fa){
LL &res=dp[u]=1;
cntv[u]=0;
rep(i,0,G[u].sz()){
int v=G[u][i].X,c=G[u][i].Y;
if(v==fa) continue;
dfs(v,u);
cntv[u]+=c;
//儿子的兄弟间的可重集排列
res=resinvfac[c]%mod;
//上下要插板
res=resget_C(cntv[v]+c-1,c-1)%mod;
//内部阶乘
res=resfac[2c]%mod;
//分步乘法
res=resdp[v]%mod;
}
//儿子的兄弟间的可重集排列
res=resfac[cntv[u]]%mod;
}

void pre(){
fac[0]=fac[1]=1;
invfac[0]=invfac[1]=1;
inv[1]=1;
rep(i,2,maxm){
fac[i]=fac[i-1]i%mod;
inv[i]=(mod-mod/i)inv[mod%i]%mod;
invfac[i]=invfac[i-1]*inv[i]%mod;
}
}

int main() {
pre();
scf("%d",&n);
for(int i=0;i<n-1;i++){
int u,v,c;
scf("%d%d%d",&u,&v,&c);
G[u].pb(mkp(v,c));
G[v].pb(mkp(u,c));
}
dfs(1,-1);
prf("%lld\n",dp[1]);
return 0;
}

//end-----------------------------------------------------------------------