POJ 1202 Family 概率,DP,高精 难度:2

http://poj.org/problem?id=1202

难度集中在输出格式上,因为输出格式所以是高精度

递推式:

血缘肯定只有从双亲传到儿子的,所以,设f,m为双亲,son为儿子,
p[i][j]为i和j之间的血缘关系,p[j][i]=p[i][j]

则:

p[son][f]=p[son][m]=0.5+0.5*p[f][m]

对于兄弟和其他不是双亲的节点j,则有

p[son][j]=0.5*(p[f][j]+p[m][j])

另外:http://swerc.up.pt/2002/Data/有数据

代码交了好几次都是RE,突然交了一次居然AC了

import java.math.BigDecimal;
import java.util.*;

class mes{
	int son,nxt;
	mes(){son=nxt=0;}
}
public class Main {
	final static int maxn=650;
	static int head[];
	static BigDecimal p[][];
	static mes s[];
	static int fa[][];
	static int n,k,m;
	static int vis[];
	static int que[];

	static void addedge(int f,int t,int ind)
	{
	    s[ind].nxt=head[f];
	    s[ind].son=t;
	    head[f]=ind;
	}
	static Scanner scanner;
	static void init()
	{
		for(int i=0;i<maxn;i++){
			head[i]=-1;
		}
		scanner = new Scanner(System.in);
		n=scanner.nextInt();
		k=scanner.nextInt();
		for(int i=0; i<k; i++)
		{
			int son,f,m;
			son=scanner.nextInt();
			f=scanner.nextInt();
			m=scanner.nextInt();
			addedge(f,son,2*i);
        	addedge(m,son,2*i+1);
        	fa[son][0]=f;
        	fa[son][1]=m;
		}
		for(int i=1; i<=n; i++)
		{
			p[i][i]=BigDecimal.ONE;
		}
	}

	static void calc()
	{
		int front=0,tail=0;
		final BigDecimal mul=BigDecimal.valueOf(0.5);
		for(int i=1; i<=n; i++)
		{
			if(fa[i][0]==0&&fa[i][1]==0&&vis[i]==0)
			{
        	   que[tail++]=i;
        	   vis[i]=1;
			}
		}
		while(tail!=front)
		{
			int tp=que[front];front++;
			vis[tp]=2;
			int f=fa[tp][0],m=fa[tp][1];
			if(f!=0&&m!=0)
			{
				p[tp][f]=p[tp][m]=p[f][tp]=p[m][tp]=(p[f][m].add(BigDecimal.ONE)).multiply(mul);
				for(int i=1; i<=n; i++)
				{
					if(i!=f&&i!=tp&&i!=m)
					{
						p[i][tp]=p[i][tp].add((p[i][m].add(p[i][f])).multiply(mul));
						p[tp][i]=p[i][tp];
					}
				}
			}

			for(int sp=head[tp]; sp!=-1; sp=s[sp].nxt)
			{
				int son=s[sp].son;
				if(vis[son]!=0)continue;
				if(vis[fa[son][0]]==2&&vis[fa[son][1]]==2)
				{
		        	que[tail++]=son;
					vis[son]=1;
				}
			}
		}
	}
	static void show()
	{
		m=scanner.nextInt();
		for(int i=0; i<m; i++)
		{
			int f,t;
			f=scanner.nextInt();
			t=scanner.nextInt();
			String ans = p[t][f].multiply(BigDecimal.valueOf(100)).toPlainString();
			if(ans.indexOf(".")>0&&ans.indexOf(".")<ans.length()){
				while(ans.length()>1&&ans.endsWith("0")){
					ans=ans.substring(0,ans.length()-1);
				}
			}
			if(ans.endsWith(".")&&ans.length()>1)ans=ans.substring(0,ans.length()-1);
			System.out.println(ans+"%");
		}
	}
	public static void main(String[] args) {
		head=new int[maxn];
		p=new BigDecimal[maxn][maxn];
		s=new mes[2*maxn];
		fa=new int[maxn][2];
		que=new int[maxn];
		vis = new int[maxn];
		for(int i=0;i<maxn;i++){p[i]=new BigDecimal[maxn];s[i]=new mes();fa[i]=new int[2];}
		for(int i=0;i<maxn;i++){for(int j=0;j<maxn;j++){p[i][j]=BigDecimal.ZERO;}}
		init();
		calc();
		show();
	}
}