题目地址:http://community.topcoder.com/stat?c=problem_statement&pm=1592&rd=4482
题目描述:在一个n x n 的棋盘上,给定一个start position,一个finish position,一个knight的移动方式,求解从start position到finish position 有多少条长度为numMoves的不同的路。
思路: ways[i][j][n]表示从start position 到 (i,j)的长度为n的路径的数量。ways[i][j][n] = ways[i-1][j][n-1]+ways[i][j-1][n-1]+ways[i-1][j-1][n-1]+ways[i-1][j+1][n-1]+ways[i] [j+1][n-1]+ways[i+1][j-1][n-1]+ways[i+1][j][n-1]+ways[i+1][j+1][n-1]+ways[i-2][j-1][n-1]+ways[i-1][j-2][n-1]+ways[i-2][j+1][n-1]+ways[i-1][j+2][n-1]+ways[i+1][j-2][n-1]+ways[i+2][j-1][n-1]+ways[i+1][j+2][n-1]+ways[i+2][j+1][n-1]
public class ChessMetric { public long howMany(int size,int[] start,int[] end,int numMoves){ long[][][] ways = new long[size][size][numMoves+1]; ways[start[0]][start[1]][0] = 1; for(int i=1;i<=numMoves;i++) for(int j=0;j<size;j++) for(int k=0;k<size;k++){ long temp=0; if(j-1>=0) temp += ways[j-1][k][i-1]; if(k-1>=0) temp += ways[j][k-1][i-1]; if(j-1>=0&&k-1>=0) temp += ways[j-1][k-1][i-1]; if(j-1>=0&&k+1<size) temp += ways[j-1][k+1][i-1]; if(k+1<size) temp += ways[j][k+1][i-1]; if(j+1<size&&k-1>=0) temp += ways[j+1][k-1][i-1]; if(j+1<size) temp += ways[j+1][k][i-1]; if(j+1<size&&k+1<size) temp += ways[j+1][k+1][i-1]; if(j-2>=0&&k-1>=0) temp += ways[j-2][k-1][i-1]; if(j-1>=0&&k-2>=0) temp += ways[j-1][k-2][i-1]; if(j-2>=0&&k+1<size) temp += ways[j-2][k+1][i-1]; if(j-1>=0&&k+2<size) temp += ways[j-1][k+2][i-1]; if(j+1<size&&k-2>=0) temp += ways[j+1][k-2][i-1]; if(j+2<size&&k-1>=0) temp += ways[j+2][k-1][i-1]; if(j+1<size&&k+2<size) temp += ways[j+1][k+2][i-1]; if(j+2<size&&k+1<size) temp += ways[j+2][k+1][i-1]; ways[j][k][i] = temp; } return ways[end[0]][end[1]][numMoves]; } }