题目
求\(n\)个点直径为\(d\)的标号树个数（多组数据）
（\(0\leq d\leq n\leq 50,n>0\)）
分析
首先特判一下\(n==d\)无解，\(d=0\)除非只有一个点，\(d=1\)除非只有两个点
然后直径很难搞，考虑把它转换成深度
设\(dp[n][h]\)表示\(n\)个点，深度不超过\(h\)的标号树个数
那么\(dp[n][h]=\sum_{i=1}^{n-1}dp[i][h-1]*dp[n-i][h]*c[n-2][i-1]*i\)
解释一下什么意思（未完待续）
代码
#include <cstdio>
#include <cstring>
#define rr register
const int mod=1000000000,N=51; long long c[N][N];
struct Big{long long p[11];}dp[N][N];
Big operator +(Big a,Big b){
rr int Len=a.p[0]>b.p[0]?a.p[0]:b.p[0],g=0;
rr Big c; memset(c.p,0,sizeof(c.p));
for (rr int i=1;i<=Len;++i){
c.p[i]=a.p[i]+b.p[i]+g;
if (c.p[i]>=mod) g=1,c.p[i]-=mod;
else g=0;
}
if (g) c.p[++Len]=g;
c.p[0]=Len;
return c;
}
Big operator -(Big a,Big b){
rr int Len=a.p[0]>b.p[0]?a.p[0]:b.p[0],g=0;
rr Big c; memset(c.p,0,sizeof(c.p));
for (rr int i=1;i<=Len;++i){
rr int s=a.p[i]-b.p[i]-g;
if (s<0) s+=mod,g=1; else g=0;
c.p[i]=s;
}
while (!c.p[Len]) --Len;
c.p[0]=Len;
return c;
}
Big operator *(Big a,Big b){
rr int Len=a.p[0]+b.p[0];
rr Big c; memset(c.p,0,sizeof(c.p));
for (int i=1;i<=a.p[0];++i)
for (int j=1;j<=b.p[0];++j)
c.p[i+j-1]+=a.p[i]*b.p[j];
for (rr int i=1;i<=Len;++i) c.p[i+1]+=c.p[i]/mod,c.p[i]%=mod;
if (!c.p[Len]) --Len;
c.p[0]=Len;
return c;
}
inline Big Num_To_Big(long long n){
rr Big c; memset(c.p,0,sizeof(c.p));
for (;n;n/=mod) c.p[++c.p[0]]=n%mod;
return c;
}
inline void zero_print(int ans){
for (rr int lim=mod/10;lim>1&&lim>ans;lim/=10) putchar(48);
}
inline void print(int ans){
if (ans>9) print(ans/10);
putchar(ans%10+48);
}
inline void Big_print(Big ans){
print(ans.p[ans.p[0]]);
for (rr int i=ans.p[0]-1;i;--i)
zero_print(ans.p[i]),print(ans.p[i]);
}
signed main(){
c[0][0]=1;
for (rr int i=1;i<N;++i){
c[i][0]=c[i][i]=1;
for (rr int j=1;j<i;++j)
c[i][j]=c[i-1][j-1]+c[i-1][j];
}
for (rr int i=0;i<N;++i) dp[1][i]=Num_To_Big(1);
for (rr int i=2;i<N;++i) dp[i][0]=Num_To_Big(0);
for (int n=2;n<N;++n)
for (int d=1;d<N;++d){
dp[n][d]=Num_To_Big(0);
if (d>=n) {dp[n][d]=dp[n][d-1]; continue;}
for (rr int i=1;i<n;++i)
dp[n][d]=dp[n][d]+Num_To_Big(c[n-2][i-1]*i)*dp[i][d-1]*dp[n-i][d];
}
for (rr int n,d;scanf("%d%d",&n,&d)==2;putchar(10)){
rr int h=d/2; rr Big ans=Num_To_Big(0);
if (n==d) {putchar(48); continue;}
if (!d) {putchar(48+(n==1)); continue;}
if (!h) {putchar(48+(n==2)); continue;}
if (d&1){
for (rr int i=h+1;i<=n;++i){
rr int j=n-i; if (j<=h) break;
ans=ans+(dp[i][h]-dp[i][h-1])*(dp[j][h]-dp[j][h-1])*Num_To_Big(c[n-2][i-1]*c[n][2]);
}
}else{
ans=(dp[n][h]-dp[n][h-1])*Num_To_Big(n);
if (h>=2){
for (rr int i=h;i<n;++i)
ans=ans-Num_To_Big(c[n-1][i]*n*i)*dp[n-i][h-1]*(dp[i][h-1]-dp[i][h-2]);
}
}
Big_print(ans);
}
return 0;
}