// GooleAnswer.cpp : Defines the entry point for the console application.
//
/*
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.
Definition
????
Class:
WordPath
Method:
countPaths
Parameters:
vector <string>, string
Returns:
int
Method 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.
Examples
0)

????
{"ABC",
 "FED",
 "GHI"}
"ABCDEFGHI"
Returns: 1
There is only one way to trace this path. Each letter is used exactly once.
1)

????
{"ABC",
 "FED",
 "GAI"}
"ABCDEA"
Returns: 2
Once we get to the 'E', we can choose one of two directions for the final 'A'.
2)

????
{"ABC",
 "DEF",
 "GHI"}
"ABCD"
Returns: 0
We can trace a path for "ABC", but there's no way to complete a path to the letter 'D'.
3)

????
{"AA",
 "AA"}
"AAAA"
Returns: 108
We 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: 56448
There are a lot of ways to trace this path.
5)

????
{"AAAAA",
 "AAAAA",
 "AAAAA",
 "AAAAA",
 "AAAAA"}
"AAAAAAAAAAA"
Returns: -1
There are well over 1,000,000,000 paths that can be traced.
6)

????
{"AB",
 "CD"}
"AA"
Returns: 0
Since 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.
*/
#include <string>
#include <vector>
#include <iostream>
#include <stdio.h>
using namespace std; //Required for TopCoder gcc compiler

//****************************************************************
//類名：WordPath
//作者：roc（txqc4@sohu.com）
//日期：2005-12-13
//用途: 本代碼為實現上述競賽題所作。
//注意事項：如欲轉載，請保持本段說明。
//****************************************************************
class WordPath
{
    typedef struct POINTtag{
        int x;
        int y;
        int count;
    }POS;
    typedef vector<POS> VETPOS;
    
public:
    
    int countPaths(vector <string> grid, string find)
    {
        int findStrLen = find.length();
        int gridSize   = grid.size();
        int gridStrLen = grid[0].length();
        
        vector <VETPOS> vec(findStrLen);
        
        int i,j,k;
        
        // 遍歷grid中的每一個字符
        for ( i = 0; i < gridSize; i ++)
        {
            for ( j= 0; j < gridStrLen; j++)
            {
                for ( k=0; k<findStrLen; k++)
                {
                    char ch =  find.at(k);
                    //如果與find中位置k的字符相等，則將相應的grid中的位置坐標保存到相應的向量中去
                    if ( ch == grid[i].at(j) )
                    {
                        POS ps;
                        ps.x =j;
                        ps.y = i;
                        
                        //位置向量0中所有坐標的初始值為1，而其他位置向量中坐標的這個字段總會被指零后才計算
                        ps.count = 1;
                        vec[k].push_back(ps);
                    }
                }
            }
        }
        
        // 如果有find中的字符在grid中不存在則返回0
        for ( k=0; k<findStrLen; k++)
        {
            if ( vec[k].size() == 0 )
                return 0;
        }
                  
        VETPOS midVes;//保存當前位置向量中符合條件點的臨時向量
        
        // 遍歷從位置1開始的位置向量
        for ( i = 1; i < findStrLen ; i ++)
        {
            midVes.clear();
            
            //遍歷當前位置向量中的所有位置坐標
            for ( j=0; j < vec[i].size(); j++)
            {
                POS cur = vec[i][j];
                
                //如果當前點與前個向量vec[i-1]中的點可以移動到達，則保存這個點到臨時變量midVes中去
                if (  pathCount(cur,vec[i-1]))
                {
                    midVes.push_back(cur);
                }
            }
            //清空原來的向量
            vec[i].clear();
            
            //如果midVes中有符合條件的點存在，則將它保存到原來的位置向量中去
            //否則返回0
            if ( midVes.size() >0 )
            {
                vec[i] = midVes;
            }
            else
            {
                return 0;
            }
        }
        
        // 統計保存在最后位置向量中的點的count值
        int count = 0;
        for ( j=0; j < vec[findStrLen-1].size(); j++)
        {
            POS cur = vec[findStrLen-1][j];
            count += cur.count;
            if (count > 1000000000 )
                return -1;
        }
        return count;
    }
    
    int pathCount(POS &ps, VETPOS& pre)
    {
        //初始為0
        ps.count = 0;
        int i;
        
        //遍歷pre中的每個位置坐標
        for ( i=0; i < pre.size(); i++)
        {
            //計算cur與pre[i]的縱橫坐標差的絕對值
            int xAbs = ps.x - pre[i].x;
            int yAbs = ps.y - pre[i].y;
            
            xAbs = xAbs<0?-xAbs:xAbs;
            yAbs = yAbs<0?-yAbs:yAbs;
           
            //判斷是否可以移動到達
            if (( xAbs == 1 && yAbs < 2 ) ||
                ( yAbs == 1 && xAbs < 2 ) )
            {
                ps.count += pre[i].count;//統計通過ps點的可能路徑數。
            }
        }
        
        return ps.count;
    }
};


//以下為測試代碼
void Load(vector <string> &grid, string &find)
{
    string array[]={"ABC",
        "FED",
        "GHI"};
    
    for (int i=0; i < sizeof(array)/sizeof(string); i++)
    {
        grid.push_back(array[i]);
        cout<<array[i]<<endl;
    }
    find = "ABCDEFGHI";
    cout <<find<<endl;
}

void Load1(vector <string> &grid, string &find)
{
    string array[]={"ABC",
 "FED",
 "GAI"};
    
    for (int i=0; i < sizeof(array)/sizeof(string); i++)
    {
        grid.push_back(array[i]);
        cout<<array[i]<<endl;
    }
    find = "ABCDEA";
    cout <<find<<endl;
}
void Load2(vector <string> &grid, string &find)
{
    string array[]={"ABABA",
 "BABAB",
 "ABABA",
 "BABAB",
 "ABABA"};
    
    for (int i=0; i < sizeof(array)/sizeof(string); i++)
    {
        grid.push_back(array[i]);
        cout<<array[i]<<endl;
    }
    find = "ABABABBA";
    cout <<find<<endl;
}
void Load3(vector <string> &grid, string &find)
{
    string array[]={"AA",
 "AA"};
    
    for (int i=0; i < sizeof(array)/sizeof(string); i++)
    {
        grid.push_back(array[i]);
        cout<<array[i]<<endl;
    }
    find = "AAAA";
    cout <<find<<endl;
}


void Load4(vector <string> &grid, string &find)
{
    string array[]={"AAAAA",
 "AAAAA",
 "AAAAA",
 "AAAAA",
 "AAAAA"};
    
    for (int i=0; i < sizeof(array)/sizeof(string); i++)
    {
        grid.push_back(array[i]);
        cout<<array[i]<<endl;
    }
    find ="AAAAAAAAAAA";
    cout <<find<<endl;
}
int main(int argc, char* argv[])
{
    WordPath word;
    vector <string> grid;
    string find;

    void(*LoadFn[])(vector <string> &grid, string &find)={
    Load,Load1,Load2,Load3,Load4};
    
    for ( int i = 0; i < sizeof(LoadFn)/sizeof(LoadFn[0]); i ++)
    {
        grid.clear();
        LoadFn[i](grid,  find);
        cout<<word.countPaths(grid,  find)<<endl;

    }

	return 0;
}