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2013
11-10

POJ 1923 Fourier’s Lines [解题报告] Java

Fourier’s Lines

问题描述 :

Joseph Fourier was a great mathematician and physicist and is well known for his mathematic series. Among all the nineteen children in his family, Joseph was the youngest and the smartest. He began to show his interest in mathematics when he was very young. After he grew up, he often corresponded with C. Bonard (a professor of mathematics at Auxerre) by exchanging letters.

In one letter written to Bonard, Fourier asked a question: how to draw 17 lines on a plane to make exactly 101 crossings, where each crossing belongs to exactly two lines. Obviously, this is an easy problem, and Figure-1 is a solution that satisfies his requirement. Now the problem for you is a universal one. Can we draw N lines on a plane to make exactly M crossings, where each crossing belongs to exactly two lines? If we can, how many pieces, at most, can these lines cut the plane into?

输入:

The input may have several sets of test data. Each set is one line containing two integers N and M (1 <= N <= 100, 0 <= M <= 10000), separated by a space. The test data is followed by a line containing two zeros, which indicates the end of input and should not be processed as a set of data.

输出:

Output one line for each set of input in the following format:

Case i: N lines cannot make exactly M crossings.

if the drawing of these lines is impossible;

or:

Case i: N lines with exactly M crossings can cut the plane into K pieces at most.

Note: Even if N or M equals to one, you should use the words “lines” and “crossings” in your output.

样例输入:

4 3
4 6
4 2
5 11
17 101
0 0

样例输出:

Case 1: 4 lines with exactly 3 crossings can cut the plane into 8 pieces at most.
Case 2: 4 lines with exactly 6 crossings can cut the plane into 11 pieces at most.
Case 3: 4 lines cannot make exactly 2 crossings.
Case 4: 5 lines cannot make exactly 11 crossings.
Case 5: 17 lines with exactly 101 crossings can cut the plane into 119 pieces at most. 

解题代码:

/* @author: */
import java.util.Scanner;
public class Main {
  
static int com(int a)
{
	return a*(a-1)/2;
}

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
   
    long line[][]=new long[101][10001];
	int i,j,k,p,q,cs=0;
    long  max,tmp;
    for(i=1;i<=100;i++)
	 line[i][0]=i+1;
    for(i=2;i<=100;i++)
	for(j=1;j<=com(i);j++)
	{
	   max=0;
	   for(k=1;k<=i-1;k++)
	   {
	     p=i-k;q=j-k*p;
	     if(q>=0)
	     {
	       if(line[p][q]!=0)
		{
		   tmp=line[p][q]+k*(p+1);
		   if(tmp>max)
		     max=tmp;
		}
	    }
	 }
	 line[i][j]=max;
       }
	while (sc.hasNext())
	{
        i=sc.nextInt();
        j=sc.nextInt();
	 if(i==0&&j==0)
		break;
	 cs++;
	if(line[i][j]>0)
      System.out.printf("Case %d: %d lines with exactly %d crossings can cut the plane into %d 
	     pieces at most.\n",cs,i,j,line[i][j]);
	else System.out.printf("Case %d: %d lines cannot make exactly %d crossings.\n",cs,i,j);
	}
   }
}

  1. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c

  2. 如果两个序列的最后字符不匹配(即X [M-1]!= Y [N-1])
    L(X [0 .. M-1],Y [0 .. N-1])= MAX(L(X [0 .. M-2],Y [0 .. N-1]),L(X [0 .. M-1],Y [0 .. N-1])
    这里写错了吧。

  3. 在方法1里面:

    //遍历所有的边,计算入度
    for(int i=0; i<V; i++)
    {
    degree = 0;
    for (j = adj .begin(); j != adj .end(); ++j)
    {
    degree[*j]++;
    }
    }

    为什么每遍历一条链表,要首先将每个链表头的顶点的入度置为0呢?
    比如顶点5,若在顶点1、2、3、4的链表中出现过顶点5,那么要增加顶点5的入度,但是在遍历顶点5的链表时,又将顶点5的入度置为0了,那之前的从顶点1234到顶点5的边不是都没了吗?

  4. 博主您好,这是一个内容十分优秀的博客,而且界面也非常漂亮。但是为什么博客的响应速度这么慢,虽然博客的主机在国外,但是我开启VPN还是经常响应很久,再者打开某些页面经常会出现数据库连接出错的提示