微软2014年4月 实习生招聘机试题 4.most frequent logs

Time Limit: 10000ms
Case Time Limit: 3000ms
Memory Limit: 256MB

Description

In a running system, there're many logs produced within a short period of time, we'd like to know the count of the most frequent logs.
Logs are produced by a few non-empty format strings, the number of logs is N(1=N=20000), the maximum length of each log is 256.
Here we consider a log same with another when their edit distance (see note) is = 5.
Also we have a) logs are all the same with each other produced by a certain format string b) format strings have edit distance  5 of each other.
Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length.
Note edit distance is the minimum number of operations (insertdeletereplace a character) required to transform one string into the other, please refer to httpen.wikipedia.orgwikiEdit_distance for more details.

Input

Multiple lines of non-empty strings.

Output

The count of the most frequent logs.

Sample In

Logging started for id:1
Module ABC has completed its job
Module XYZ has completed its job
Logging started for id:10
Module ? has completed its job

Sample Out

3

 

以下为个人给出的一个解，计算编辑距离使用了动态规划

#include <iostream>#include <fstream>#include <string>using namespace std;//计算编辑距离，使用动态规划法int calEditDis(char* a,char* b){int len_a=strlen(a);int len_b=strlen(b);int tmp=0;int leftUp,left,up;len_a++; //维数分配 需有边界1维len_b++;int result = 0;int* assistArray; //二维数组 assistArray[len_a][len_b]assistArray = new int [(len_a)*(len_b)]();for(int i=0;i<len_a;i++)assistArray[i*len_b]=i; //assistArray[i][0]for(int j=0;j<len_b;j++)assistArray[j] = j;     //assistArray[0][j]//cout<<"********************************************"<<endl;for(int i=1;i<len_a;i++){for(int j=1;j<len_b;j++){if(a[i-1]==b[j-1])leftUp = assistArray[(i-1)*(len_b)+(j-1)];else leftUp = assistArray[(i-1)*(len_b)+(j-1)] + 1;left = assistArray[i*len_b+(j-1)] + 1;up   = assistArray[(i-1)*len_b+j] + 1 ;assistArray[i*len_b+j] = left<up?(left<leftUp?left:leftUp):(up<leftUp?up:leftUp);//cout<<assistArray[i*len_b+j]<<" ";}//cout<<endl;}//cout<<"**********************************************"<<endl;result = assistArray[len_a*len_b-1];delete[] assistArray;return result;}#define LOGNUM (20000)#define LOGLEN (256+1)typedef struct RECORDER{bool useful;int  index;int  num;}RECORDER;int main(){char (*ptr)[LOGLEN];ptr = new char[LOGNUM][LOGLEN];int dis=0;bool newLog =false;int result = 0;//record[LOGNUM] 数组中记录了 出现次数 依次从多到少的相关信息，在整个过程中都维护该数据结构RECORDER record[LOGNUM]; int count = 0;memset(record,0,sizeof(record));fstream fstr("testCase.txt");for(int i= 0,j=0,k=0;i< LOGNUM;i++){newLog = true;if(fstr.eof())break;fstr.getline(ptr[i],LOGLEN);if(strlen(ptr[i])==0)continue;for(j=0;j<count;j++){dis = calEditDis(ptr[(record[j].index)],ptr[i]);if(dis <=5){record[j].num++;newLog = false;break;}}if(newLog == true){count++;record[count-1].index=i;record[count-1].useful = true;record[count-1].num = 1;}else if(j!=0){if(j>=1 && record[j].num > record[j-1].num){k=j;while(k>=1 && record[k].num > record[k-1].num)k--;{RECORDER tmp = record[j];record[j] = record[k];record[k] = tmp;}}}}cout<<record[0].num<<endl;cout<<ptr[record[0].index]<<endl;delete[] ptr;}