The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 515    Accepted Submission(s): 146

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The stringS is dominated by T if S is a substring of T.

 

Input

The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.

Output

For each test case, output a dominator if exist, or No if not.

Sample Input

310youbetterworsericherpoorersicknesshealthdeathfaithfulnessyoubemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness5abccdeabcdeabcdebcde3aaaaaaaaabaaaac

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulnessabcdeNo

Source

2017 ACM/ICPC Asia Regional Qingdao Online 

唉 ，，，，好坑啊，开始用ac自动机，要不超内存，要不就超时，挑了将近两三个小时，没ac，比赛结束，听说用kmp就能过，然后试了下，AC

代码如下：

#include <iostream>  #include <cstdio>  #include <stdlib.h>#include <cstring>  #include <queue>  using namespace std;int kmp_find(const char* target,const char* pattern){    const int target_length = strlen(target);    const int pattern_length = strlen(pattern);    int * overlay_value = new int[pattern_length];    overlay_value[0] = -1;//存前缀和后缀相同元素的长度    int index = 0;    for(int i=1;i<pattern_length;++i)    {        index = overlay_value[i-1];        while(index>=0 && pattern[index+1]!=pattern[i])        {            index  = overlay_value[index];        }        if(pattern[index+1]==pattern[i])        {            overlay_value[i] = index +1;        }        else        {            overlay_value[i] = -1;        }    }    //match algorithm start    int pattern_index = 0;    int target_index = 0;    while(pattern_index<pattern_length&&target_index<target_length)    {        if(target[target_index]==pattern[pattern_index])        {            ++target_index;            ++pattern_index;        }        else if(pattern_index==0)        {            ++target_index;        }        else        {            pattern_index = overlay_value[pattern_index-1]+1;        }    }    delete [] overlay_value;    if(pattern_index==pattern_length)    {        return target_index-pattern_index;    }    else    {        return -1;    }   }  char s[200002];char* p[100005];int main()  {      int t,n,m,len,i;     scanf("%d",&t);      while(t--)      {      scanf("%d",&n);    memset(s,'\0',sizeof(s));    len=m=0;    p[0]=s;    for(int i=0;i<n;i++)    {    scanf("%s",p[i]);    if(strlen(p[i])>len){len=strlen(p[i]);m=i;}     p[i+1]=p[i]+strlen(p[i])+1; }        int ans=0;for(int i=0;i<n;i++){if(i!=m&&kmp_find(p[m],p[i])!=-1)ans++;}if(ans==n-1)printf("%s\n",p[m]);elseprintf("No\n");    }      return 0;  }