bzoj 1005: [HNOI2008]明明的煩惱

題目鏈接：http://www.lydsy.com:808/JudgeOnline/problem.php?id=1005

題目大意：給出標號爲1到N的點,以及某些點最終的度數,允許在任意兩點間連線,可產生多少棵度數滿足要求的樹。

題目分析：http://www.cnblogs.com/zhj5chengfeng/p/3278557.html

代碼參考：

import java.util.*;
import java.io.BufferedInputStream;
import java.math.*;
import java.lang.*;
public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(new BufferedInputStream(System.in));
		BigInteger f[] = new BigInteger [2000];
		int d[] = new int [20000];
		int i, n, cnt, sum;
		f[0] = BigInteger.ONE;//計算階乘
		for(i=1; i<=1000; ++i) f[i] = f[i-1].multiply(BigInteger.valueOf(i));
			n = in.nextInt();
			cnt = sum = 0;
			boolean flag = false;
			for(i=0; i<n; ++i) {
				d[i] = in.nextInt();
				if(d[i] == 0 || d[i] > n-1) flag = true;//特殊情況判斷
				if(d[i] == -1) continue;//只計算度數有限定的點
				sum += d[i] - 1;
				cnt++;//度數有限定的點的個數
			}
			//特殊情況的處理
			if(flag == true) {
				System.out.println("0");
				return;
			}
			if(n == 1) {
				if(d[0] == -1 || d[0] == 0) System.out.println("1");
				else System.out.println("0");
				return;
			}
			if(n == 2) {
				if((d[0] == -1 || d[0] == 0) && (d[1] == -1 || d[1] == -1)) 
					System.out.println("1");
				else System.out.println("0");
				return;
			}
			//根據公式計算
			BigInteger ans = f[n-2];
			ans = ans.divide(f[n-2-sum]);
			for(i=0; i<n; ++i) {
				if(d[i] == -1) continue;
				ans = ans.divide(f[d[i]-1]);
			}
			BigInteger tmp = BigInteger.valueOf(n-cnt).pow(n-2-sum);
			ans = ans.multiply(tmp);
			System.out.println(ans);
		}
	}