一道 Google 竞赛题的解法

一道 Google 竞赛题的解法

作者：roc

下载源代码

　　本人于2005年12月13日凌晨参加了google中国编程挑战赛的入围阶段的赛事。虽然最终我感觉自己做出了这道级别为high到mid间的赛题，但是却发现那时入围赛事早已经结束了......
　　相信 vckbase 中的不少朋友肯定也参加了那场入围赛，所以我打算把自己的解法写出来，一则虽然题目中的测试用例是全部通过了，但这并不能保证我的解法是正确的，希望大家批评指教；二则相信其他朋友也一定有更好的解法大家一起讨论讨论。希望，这篇文章能起到抛砖引玉的效果。

一、竞赛题目

Problem Statement You are given a String[] grid representing a rectangular grid of letters. You are also given a String find, a word you are to find within the grid. The starting point may be anywhere in the grid. The path may move up, down, left, right, or diagonally from one letter to the next, and may use letters in the grid more than once, but you may not stay on the same cell twice in a row (see example 6 for clarification).You are to return an int indicating the number of ways find can be found within the grid. If the result is more than 1,000,000,000, return -1.DefinitionClass:WordPathMethod:countPathsParameters:vector < string >, stringReturns:intMethod signature:int countPaths(vector < string> grid, string find)(be sure your method is public)Constraints-grid will contain between 1 and 50 elements, inclusive.-Each element of grid will contain between 1 and 50 uppercase (''A''-''Z'') letters, inclusive.-Each element of grid will contain the same number of characters.-find will contain between 1 and 50 uppercase (''A''-''Z'') letters, inclusive.Examples0){"ABC","FED","GHI"}"ABCDEFGHI"Returns: 1There is only one way to trace this path. Each letter is used exactly once.1){"ABC","FED","GAI"}"ABCDEA"Returns: 2Once we get to the ''E'', we can choose one of two directions for the final ''A''.2){"ABC","DEF","GHI"}"ABCD"Returns: 0We can trace a path for "ABC", but there''s no way to complete a path to the letter ''D''.3){"AA","AA"}"AAAA"Returns: 108We can start from any of the four locations. From each location, we can then move in any of the three possible directions for our second letter, and again for the third and fourth letter. 4 * 3 * 3 * 3 = 108.4){"ABABA","BABAB","ABABA","BABAB","ABABA"}"ABABABBA"Returns: 56448There are a lot of ways to trace this path.5){"AAAAA","AAAAA","AAAAA","AAAAA","AAAAA"}"AAAAAAAAAAA"Returns: -1There are well over 1,000,000,000 paths that can be traced.6){"AB","CD"}"AA"Returns: 0Since we can''t stay on the same cell, we can''t trace the path at all.This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

int countPaths(vector < string grid, string find)

的函数，grid相当于一个字母矩阵，grid中的每个字符串含有相同个数的字母，这些字母都是大写的字母,从''A''到''Z''，grid中字母个数的范围是1-50。参数find是要求你在grid中搜索路径的字符串，其同样只含有''A''到''Z''的字符，其个数范围同样是1-50。搜索起点可以从grid中的任何一点开始，然后可以向上，向下，向左，向右，以及对角线移动一格。grid中的每个位置的字母可以多次使用。但路径不能在相同位置停留两次(见用例6)。返回值是个整型数据，表示搜索到的路径总数数。如果这个数值大于1,000,000,000, 则返回-1。

二、我的解题步骤

第一步.使用人脑按照题目要求，实际计算一遍。

　　这里先用例1为例。在例1的find字符串的第一个字符是''A'',我们可以看到在相应的grid中，''A''在两个位置上出现，由于起点不限，所以我们搜索到的起点有2个；find的第2个字符是B，我们可以看到grid中只出现了1个B,而按照题目要求移动的话，显然只有左上角的那个A可以移动到这个B的位置。我们以次类推一值移动到E,都是唯一一条路经，但最后一个字符还是A,这时按照题目要求，2个A都可以从E移动到达，并且题目也说明每个位置可以多次移动停留，所以这时2个A都符合条件。