Google编程大赛入围赛750分真题

Google编程大赛入围赛750分真题 第五组

思路：

广度搜索。

数据结构： 两个vector，一个保存当前的位置 current，一个保存下一步的位置 next

算法:
1. 在矩阵里找字符串的第一个字母，然后放到current里
2. 逐个搜寻current里的字符的邻居，看是不是字符串里的第二个字符，是就放到next里面
3. 判断next的长度是不是超过1000000
4. 将next赋给current，将next清空
5. 重复第一步

Problem Statement
牋牋
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
牋牋
Class:
WordPath
Method:
countPaths
Parameters:
String[], String
Returns:
int
Method signature:
int countPaths(String[] grid, String find)
(be sure your method is public)
牋牋

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)

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

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

牋牋
{"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)

牋牋
{"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)

牋牋
{"ABABA",
"BABAB",
"ABABA",
"BABAB",
"ABABA"}
"ABABABBA"
Returns: 56448
There are a lot of ways to trace this path.
5)

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

牋牋
{"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.

我的思路如下，不过运行未能返回正确结果，仅供参考：
using System;
using System.Collections;
public class WordPath
{
public int countPaths(string[] grid, string find)
{
int m = 0;
int h = grid.Length;

for(int i = 0; i < h ; i ++)
{
for(int j = 0; j < grid[i].Length ; j ++)
{
if(grid[i][j] == find[0])
{
// all
m += GetNext(grid,find,i,j,1,m);
}
}

}
return m;
}

private int GetNext(string[] grid,string find , int m ,int n ,int f,int t)
{
int x = grid.Length;
int[,] im = {{0,1},{0,-1},{1,0},{-1,0},{-1,1},{1,-1},{-1,-1},{1,1}};



for(int i = 0; i < 8; i++)
{
int p = im[i,0]+m;
int q = im[i,1]+n;
if(p>-1 && q>-1 && p<x && q<x && grid[p][q] == find[f])
{
if(f+1 == find.Length)
{
t += 1;
}
else
GetNext(grid,find,p,q,f+1,t);
}
}

return t;
}
}
public class WordPath {
private String find="";
private static int count=0;
public WordPath() {
}


public int countPaths(String[] grid, String find){
this.find=find;
char[][] cs=new char[grid.length][grid[0].length()];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j <grid[i] .length(); j++) {
cs[i][j]=grid[i].charAt(j);
}
}

for (int i = 0; i < cs.length; i++) {
for (int j = 0; j <cs[i] .length; j++) {
if(cs[i][j]==find.charAt(0))
{
go(cs,0,i,j);
}
}
}




if(count>1000000000)
count=-1;

return count;
}
public void go(char[][] cs,int index,int x,int y)
{
if(index==find.length()-1)
{
count++;
return ;
}
for (int i = -1; i < 2; i++) {
for (int j = -1; j < 2; j++) {
if(i==0&&j==0)
continue;
if(x+i<0||x+i>cs.length-1)
continue ;
if(y+j<0||y+j>cs[0].length-1)
continue ;
if(cs[x+i][y+j]==find.charAt(index+1))
go(cs,index+1,x+i,y+j);

}
}


}

public static void main(String[] args) {
WordPath wordPath1 = new WordPath();

}

}

namespace S3{

const unsigned int MAX_COUNT = 1000000000;
const unsigned int MAX_UINT = 0xFFFFFFFF;

struct cell
{
char ch;
int count[2];
};

class CGrid
{
typedef cell* LPCELL;

public:
CGrid() : m_RowCount(0), m_ColCount(0), m_ppCells(NULL) { }
~CGrid() { Clear(); }

bool Initialize(const std::vector<std::string>& grid)
{
Clear();
m_RowCount = grid.size();
if (m_RowCount <= 0)
{
return false;
}

m_ColCount = grid[0].size();
if (m_ColCount <= 0)
{
return false;
}

for (int i=0; i<m_RowCount; i++)
{
if (grid[i].size() != m_ColCount)
{
return false;
}
}

int nCount = m_RowCount * m_ColCount;
m_ppCells = new LPCELL[nCount];
if (m_ppCells == NULL)
{
return false;
}

for (i=0; i<m_RowCount; i++)
{
for (int j=0; j<m_ColCount; j++)
{
LPCELL p = new cell;
m_ppCells[i * m_ColCount + j] = p;
p->ch = grid[i][j];
}
}

return true;
}

int CountPath(const std::string& find)
{
int nLen = find.size();
if (nLen <= 0)
{
return 0;
}

con

#---------------in python-----------------------

### example 0
### Returns: 1
##grid = ['ABC', 'FED', 'GHI']
##find = 'ABCDEFGHI'

### example 1
### Returns: 2
##grid = ['ABC', 'FED', 'GAI']
##find = 'ABCDEA'

### example 2
### Returns: 0
##grid = ['ABC', 'DEF', 'GHI']
##find = 'ABCD'

### example 3
### Returns: 108
##grid = ['AA', 'AA']
##find = 'AAAA'

### example 4
### Returns: 56448
##grid = ['ABABA', 'BABAB', 'ABABA', 'BABAB', 'ABABA']
##find = 'ABABABBA'

### example 5
### Returns: -1
grid = ['AAAAA', 'AAAAA', 'AAAAA', 'AAAAA', 'AAAAA']
find = 'AAAAAAAAAAA'

### example 6
### Returns: 0
##grid = ['AB', 'CD']
##find = 'AA'

class WordPath:
def __init__(self):
pass
def countPaths(self, grid, find):

#prepare the searchResult matrix
self.searchResult = []
for i in range(len(grid)):
self.searchResult.append([])
for i in range(len(grid)):
for j in range(len(grid[0])):
self.searchResult[i].append({})

count = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
result = self.search(grid, i, j, find)
if result == -1:
return -1
count = count + result
感觉这是最好的答案：
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <time.h>
using namespace std;

const int neigh[8][2] = {{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};
const int Max = 1000000000;
class WordPath
{
private:
int result ;
int map[3][52][52];
int X,Y;
bool level;
public:
inline int fresult()
{
int i,j;
result = 0;
for (i=1; i <= X; i++)
for (j=1; j <= Y; j++)
{
result += map[level][i][j];
if (result > Max) return -1;
}
return result;
}
int countPaths(vector <string> grid, string findpath)
{
int i,j,k,l,m,n,q;
X = grid[0].size();
Y = X;
memset(map,0,sizeof map);
level = 0;
for (i=1; i <= X; i++)
for (j=1; j <= Y; j++)
{
map[2][i][j] = grid[i-1][j-1];
if (map[2][i][j] == findpath[0])
map[level][i][j] = 1;
};
if (findpath.length()==1)
return fresult();

for (k=1; k < findpath.length(); k++)
{
level = !level;
for (i=1; i <= X; i++)
for (j=1; j <= Y; j++)
if (map[2][i][j] == findpath[k])
for (l=0; l < 8; l++)
{
map[level][i][j] += map[!level][i+neigh[l][0]][j+neigh[l][1]];
if (ma

算法可以，远比想的简单。

//--------------------------------------
import java.util.*;
import java.text.*;

class Data {
private static char[][] rectangle = {
{'A', 'B','A','B', 'A'},
{'B', 'A','B','A', 'B'},
{'A', 'B','A','B', 'A'},
{'B', 'A','B','A', 'B'},
{'A', 'B','A','B', 'A'},
};
private static String stf = "ABABABBA";
public static String stringToFind() {
return stf;
}
public static char[][] rectangleToSearch() {
return rectangle;
}
public int MAXIMUM = 100000;
}

class Position {
private int x, y;
private static int recWidth, recHeight;
private static Data d;

public static void Init(Data d_) {
d = d_;
recWidth = d.rectangleToSearch().length;
recHeight = d.rectangleToSearch()[0].length;
}

public Position(int x, int y) {
this.x = x;
this.y = y;
}

public int X() {
return x;
}
public int Y() {
return y;
}
public char charValue() {
return d.rectangleToSearch()[x][y];
}
public ArrayList neighbor() {
ArrayList result = new ArrayList();
int nx, ny;
for ( int i = -1; i != 2