解题：HEOI 2013 SAO

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int N=1005,mod=1e9+7;
 6 int fac[N],inv[N],siz[N],f[N][N];
 7 int p[N],noww[2*N],goal[2*N],val[2*N];
 8 int T,n,t1,t2,cnt; char rd[5];
 9 void Link(int f,int t,int v)
10 {
11     noww[++cnt]=p[f],p[f]=cnt;
12     goal[cnt]=t,val[cnt]=v;
13 }
14 int Qpow(int x,int k)
15 {
16     if(k==1) return x;
17     int tmp=Qpow(x,k/2);
18     return k%2?1ll*tmp*tmp%mod*x%mod:1ll*tmp*tmp%mod;
19 }
20 int C(int a,int b)
21 {
22     if(a<b) return 0;
23     return 1ll*fac[a]*inv[b]%mod*inv[a-b]%mod;
24 }
25 void Prework()
26 {
27     fac[0]=inv[0]=1;
28     for(int i=1;i<=1000;i++) 
29         fac[i]=1ll*fac[i-1]*i%mod;
30     inv[1000]=Qpow(fac[1000],mod-2);
31     for(int i=999;i;i--)
32         inv[i]=1ll*inv[i+1]*(i+1)%mod;
33     scanf("%d",&T);
34 }
35 void Init()
36 {
37     memset(p,0,sizeof p);
38     memset(f,0,sizeof f);
39     cnt=0,scanf("%d",&n);
40     for(int i=1;i<n;i++)
41     {
42         scanf("%d%s%d",&t1,rd,&t2),t1++,t2++;
43         Link(t1,t2,rd[0]==‘<‘),Link(t2,t1,rd[0]==‘>‘);
44     }
45     for(int i=1;i<=n;i++) f[i][1]=1;
46 }
47 void DFS(int nde,int fth)
48 {
49     siz[nde]=1;
50     for(int i=p[nde];i;i=noww[i])
51     {
52         int g=goal[i];
53         if(g!=fth)
54         {
55             DFS(g,nde);
56             if(val[i])
57             {
58                 for(int j=siz[nde]+siz[g];j;j--)
59                 {
60                     int tmp=0;
61                     for(int k=1;k<=min(j,siz[nde]);k++)
62                         if(j-k+1<=siz[g])
63                             tmp+=1ll*f[nde][k]*(f[g][siz[g]]-f[g][j-k]+mod)%mod*C(j-1,k-1)%mod*C(siz[nde]+siz[g]-j,siz[nde]-k)%mod,tmp%=mod;
64                     f[nde][j]=tmp;
65                 }
66             }
67             else
68             {
69                 for(int j=siz[nde]+siz[g];j;j--)
70                 {
71                     int tmp=0;
72                     for(int k=1;k<=min(j,siz[nde]);k++)
73                         if(min(j-k,siz[g])>=1)
74                             tmp+=1ll*f[nde][k]*f[g][min(j-k,siz[g])]%mod*C(j-1,k-1)%mod*C(siz[nde]+siz[g]-j,siz[nde]-k)%mod,tmp%=mod;
75                     f[nde][j]=tmp;
76                 }
77             }
78             siz[nde]+=siz[g];
79         }
80     }
81     for(int i=1;i<=siz[nde];i++)
82         f[nde][i]=(f[nde][i]+f[nde][i-1])%mod;
83 }
84 int main()
85 {
86     Prework();
87     while(T--)
88         Init(),DFS(1,0),printf("%d\n",f[1][n]);
89     return 0;
90 }