codeforces Gym 100338H High Speed Trains （递推，高精度）

递推就好了，用二项式定理算出所有连边的方案数，减去不合法的方案，

每次选出一个孤立点，那么对应方案数就是上次的答案。

枚举选几个孤立点和选哪些，选到n-1个点的时候相当于都不选，只减1。

要用到高精度，直接开100*100的组合数数组会MLE，用滚动数组优化一下就好了。

不会java，python太伤了

#include<bits/stdc++.h>
using namespace std;

const int MAXN = 20000;
struct bign
{
    int len, s[MAXN];
    bign ()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    bign (int num) { *this = num; }
    bign (const char *num) { *this = num; }
    bign operator = (const int num)
    {
        char s[MAXN];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator = (const char *num)
    {
        for(int i = 0; num[i] == '0'; num++) ;
        len = strlen(num);
        for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
    bign operator + (const bign &b) const
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; g || i < max(len, b.len); i++)
        {
            int x = g;
            if(i < len) x += s[i];
            if(i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    bign operator += (const bign &b)
    {
        *this = *this + b;
        return *this;
    }
    void clean()
    {
        while(len > 1 && !s[len-1]) len--;
    }
    bign operator * (const bign &b)
    {
        bign c;
        c.len = len + b.len;
        for(int i = 0; i < len; i++)
        {
            for(int j = 0; j < b.len; j++)
            {
                c.s[i+j] += s[i] * b.s[j];
            }
        }
        for(int i = 0; i < c.len; i++)
        {
            c.s[i+1] += c.s[i]/10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    }
    bign operator *= (const bign &b)
    {
        *this = *this * b;
        return *this;
    }
    bign operator - (const bign &b)
    {
        bign c;
        c.len = 0;
        for(int i = 0, g = 0; i < len; i++)
        {
            int x = s[i] - g;
            if(i < b.len) x -= b.s[i];
            if(x >= 0) g = 0;
            else
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    bign operator -= (const bign &b)
    {
        *this = *this - b;
        return *this;
    }
    bign operator / (const bign &b)
    {
        bign c, f = 0;
        for(int i = len-1; i >= 0; i--)
        {
            f = f*10;
            f.s[0] = s[i];
            while(f >= b)
            {
                f -= b;
                c.s[i]++;
            }
        }
        c.len = len;
        c.clean();
        return c;
    }
    bign operator /= (const bign &b)
    {
        *this  = *this / b;
        return *this;
    }
    bign operator % (const bign &b)
    {
        bign r = *this / b;
        r = *this - r*b;
        return r;
    }
    bign operator %= (const bign &b)
    {
        *this = *this % b;
        return *this;
    }
    bool operator < (const bign &b)
    {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] < b.s[i];
        }
        return false;
    }
    bool operator > (const bign &b)
    {
        if(len != b.len) return len > b.len;
        for(int i = len-1; i >= 0; i--)
        {
            if(s[i] != b.s[i]) return s[i] > b.s[i];
        }
        return false;
    }
    bool operator == (const bign &b)
    {
        return !(*this > b) && !(*this < b);
    }
    bool operator != (const bign &b)
    {
        return !(*this == b);
    }
    bool operator <= (const bign &b)
    {
        return *this < b || *this == b;
    }
    bool operator >= (const bign &b)
    {
        return *this > b || *this == b;
    }
    string str() const
    {
        string res = "";
        for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
        return res;
    }
};
istream& operator >> (istream &in, bign &x)
{
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream &out, const bign &x)
{
    out << x.str();
    return out;
}

bign fpow(bign a,int b)
{
    bign ret = 1;
    while(b){
        if(b&1) ret *= a;
        a *= a;
        b >>= 1;
    }
    return ret;
}
const int maxn = 101;
bign tab[maxn];

//typedef long long ll;
bign C[2][maxn];


int main()
{
    freopen("trains.in","r",stdin);
    freopen("trains.out","w",stdout);
    tab[1] = 0;
    int n;
    scanf("%d",&n);
    C[2&1][1]  = C[2&1][0] = 1;
    for(int i = 2; i <= n; i++){
        bign Full = fpow(2,i*(i-1)/2);
        int pre = i&1,cur = pre^1;
        C[cur][0] = 1;
        for(int j = 1; j <= i; j++) C[cur][j] = C[pre][j-1] + C[pre][j];

        for(int j = 2; j < i; j++){
            Full -= tab[j]*C[cur][j];
        }
        tab[i] = Full - 1;
    }
    cout<<tab[n]<<endl;
    return 0;
}