hdu Mahjong tree 树dp

题意：给出一棵有根树编号1~n，1为根，树上放n个麻将，编号1~n。要求是对于每个节点放置的麻将编号，这颗树的子节点放置编号必须连续，每一颗子树的放置编号也必须连续，最后求方案数%mod（1e9+7）

通过画一下就可以知道，对于每个节点来说，设它的子树大小（节点数量）为sz，那么sz>=2的子树个数不能超过2，否则方案数则一定为0.这样的话再处理一下sz为1的子树数量，最后统计一下每个节点的方案数即可

//#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef pair <int, int> pl;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 100010;
const ll mod = 1e9 + 7;

struct node{
    int v, nxt;
}e[N<<1];

ll fac[N];
int head[N];
int n, cnt;
int dp[N];
int sz[N];

void init()
{
    memset( head, -1, sizeof( head ) );
    cnt = 0;
}

void init_fac()
{
    fac[0] = 1;
    for( int i = 1; i <= N-10; ++i )
        fac[i] = fac[i-1] * i % mod;
}

void add( int u, int v )
{
    e[cnt].v = v;
    e[cnt].nxt = head[u];
    head[u] = cnt++;

    e[cnt].v = u;
    e[cnt].nxt = head[v];
    head[v] = cnt++;
}

void dfs( int u, int fa )
{
    int x = 0, y = 0;
    dp[u] = sz[u] = 1;
    ll cur = 1;
    for( int i = head[u]; ~i; i = e[i].nxt ) {
        int v = e[i].v;
        if( v == fa )
            continue;
        dfs( v, u );
        sz[u] += sz[v];
        cur = cur * dp[v] % mod;
        if( sz[v] == 1 )
            x++;
        else {
            y++;
            if( y >= 3 ) {
                dp[u] = 0;
                return;
            }
        }
    }
    if( y == 0 && x == 0 )
        return;
    if( y == 0 ) {
        dp[u] = cur * fac[x] % mod;
        return;
    }
    else {
        dp[u] = cur * 2 % mod * fac[x] % mod;
        return;
    }
}

int main()
{
    init_fac();
    int tot;
    scanf("%d", &tot);
    for( int ca = 1; ca <= tot; ++ca ) {
        init();
        scanf("%d", &n);
        for( int i = 1, u, v; i < n; ++i ) {
            scanf("%d%d", &u, &v);
            add( u, v );
        }
        printf("Case #%d: ", ca);
        if( n == 1 ) {
            printf("1\n");
            continue;
        }
        dfs( 1, 1 );
        printf("%d\n", dp[1] * 2 % mod);
    }
    return 0;
}