String Statistics(2008年珠海市ACM程序设计竞赛)

String Statistics

时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte

描述

 

You have an n × n matrix of lower-case letters. You start from somewhere in the matrix, walk in one of the eight directions (left, right, up, down, up-left, up-right, down-left, down-right) and concatenate every letter you meet, in the order you see them, to form a string. You can stop anywhere (possibly not moving at all!), but you cannot go out of the matrix, or change direction in the middle of the journey.How many different non-empty strings can you get?

 

输入

 

The first line contains t (1 ≤ t ≤ 10), the number of test cases followed. Each test case begins with one integer n(1 ≤ n ≤ 30), followed by n lines, containing the matrix. The matrix will contain lower-case letters only.

 

输出

 

For each test case, print the number of different strings you can get.

 

样例输入

2

2

aa

bb

3

aba

bcc

daa

 

样例输出

6

31

 

题目大意：给出一个n*n的矩阵，有上，下，左，右，左上，左下，右上，右下八个方向，从任意位置开始，按同一方向走动，可以在任意位置停止，输出总共有多少种不为空且各不相同的字符串。

题解：固定方向，用set容器来插入字符串，这样可以去重

#include<iostream>
#include<string>
#include<set>
using namespace std;

int dx[8]={0,0,1,-1,1,1,-1,-1}; //八个方向
int dy[8]={1,-1,0,0,-1,1,-1,1};
int n;
char op[40][40];
string str;    
set<string>p;

void dfs(int i,int j,int k) //按同方向搜索
{
    if(i>0&&j>0&&i<=n&&j<=n)
    {
        str+=op[i][j];
        p.insert(str);
        dfs(i+dx[k],j+dy[k],k);
    }
}
int main()
{
    int i,j,k,t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        getchar();
        for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        cin>>op[i][j];
        p.clear(); //每寻找一次需要清除容器里的东西        
 　　　　for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        for(k=0;k<8;k++)
        {
        　　str = "";
       　　 str+=op[i][j];
       　　 p.insert(str);
       　　 dfs(i+dx[k],j+dy[k],k);
        }
        cout<<p.size()<<endl;
    }
    return 0;
}