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2013
11-11

POJ 2704 Pascal’s Travels [解题报告] Java

Pascal’s Travels

问题描述 :

An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress.

Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed.

Figure 1 Figure 2

输入:

The input contains data for one to thirty boards, followed by a final line containing only the integer -1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 0-9, with no spaces between them.

输出:

The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 263 paths for any board.

样例输入:

4
2331
1213
1231
3110
4
3332
1213
1232
2120
5
11101
01111
11111
11101
11101
-1

样例输出:

3
0
7

温馨提示:

Brute force methods examining every path will likely exceed the allotted time limit. 64-bit integer values are available as long values in Java or long long values using the contest’s C/C++ compilers.

解题代码:

//* @author: [email protected]
import java.io.*;
public class Main
{
 static int[][] p;
 static int a;
 static long[][] w;
 public static void main(String[] args) throws IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  while(true)
  {
    a=Integer.parseInt(in.readLine());
    if(a==-1) break;
    p=new int[a][a];
    w=new long[a][a];
    w[0][0]=1;
    for(int i=0;i< a; i++)
    {
	for(int j=0;j< a; j++)
		p[i][j]=in.read()-'0';
	in.readLine();
    }
    for(int j=0;j< a;j++)
	if(p[0][j]+j< a) w[0][p[0][j]+j]+=w[0][j];
    for(int i=1;i< a;i++)
    {
	for(int j=0;j< a;j++)
		if(p[i-1][j]+i-1< a) w[p[i-1][j]+i-1][j]+=w[i-1][j];
	for(int j=0;j< a-1;j++)
		if(p[i][j]+j< a) w[i][p[i][j]+j]+=w[i][j];	
     }
     System.out.println(w[a-1][a-1]);
   }
  }
}

  1. 第二种想法,我想来好久,为啥需要一个newhead,发现是把最后一个节点一直返回到嘴上面这层函数。厉害,这道题之前没样子想过。