TopCoder-ChessMetric

来源：互联网发布：python量化投资入门编辑：程序博客网时间：2024/05/29 11:24

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];}}

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