String Statistics

時間限制(普通/Java):1000MS/3000MS????????? 運行內存限制:65536KByte

描述

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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?

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輸入

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The ?rst 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.

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輸出

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

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樣例輸入

2 2 aa bb 3 aba bcc daa

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樣例輸出

6 31 題目大意：給出一個n*n的矩陣，有上，下，左，右，左上，左下，右上，右下八個方向，從任意位置開始，按同一方向走動，可以在任意位置停止，輸出總共有多少種不為空且各不相同的字符串。 題解：固定方向，用set容器來插入字符串，這樣可以去重 #include<iostream> #include<string> #include<set> using namespace std;int dx[8]={0,0,1,-1,1,1,-1,-1}; //八個方向
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(); //每尋找一次需要清除容器里的東西
　　　　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; }

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轉載于:https://www.cnblogs.com/lavender913/p/3319857.html

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