Codeforces Gym 100338H High Speed Trains 组合数学+dp+高精度

原题链接：http://codeforces.com/gym/100338/attachments/download/2136/20062007-winter-petrozavodsk-camp-andrew-stankevich-contest-22-asc-22-en.pdf

题意

给你n个点，让你连边，使得每个点的度至少为1，问你方案数。

题解

从正面考虑非常困难，应从反面考虑，取 i 点出来不连，这样的取法一共有C(n,i)种取法，其他的连好，而这样就是子问题了。那么dp[n]=2^(n*(n-1)/2)-sum(C(n,i)*dp[n-i])-1，2<=i<=n-1

代码

import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;

public class Main {
    static int n;
    static BigInteger dp[] = new BigInteger[110];
    static BigInteger c[][] = new BigInteger[110][110];
    static BigInteger ps[] = new BigInteger[10010];
    static BigInteger x;
    public Main() throws FileNotFoundException{
        Scanner cin = new Scanner(new File("trains.in"));
        PrintWriter cout = new PrintWriter(new File("trains.out"));
        n = cin.nextInt();
        //System.out.println("n = " + n);
        for(int i=1;i<=100;i++){
            c[i][0] = new BigInteger("1");
            c[i][1] = new BigInteger(Integer.toString(i));
            c[i][i] = new BigInteger("1");
        }
        for(int i=2;i<=100;i++){
            for(int j=2;j<i;j++){
                c[i][j] = c[i-1][j].add(c[i-1][j-1]);
            }
        }
        ps[0] = new BigInteger("1");
        for(int i=1;i<=10001;i++){
            ps[i] = ps[i-1].multiply(new BigInteger("2"));
        }
        dp[2] = new BigInteger("1");
        dp[3] = new BigInteger("4");
        for(int i=4;i<=n;i++){
            x = new BigInteger("0");
            for(int j=2;j<=i-1;j++){
                x = x.add(c[i][i-j].multiply(dp[j]));
            }
            x = x.add(new BigInteger("1"));
            dp[i] = ps[i*(i-1)/2].subtract(x);
        }
        cout.println(dp[n]);
        //System.out.println(dp[n]);
        cin.close();
        cout.close();
    }

    public static void main(String[] args) throws FileNotFoundException {
        new Main();
    }
}