KMP算法

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The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2474 Accepted Submission(s): 888

Problem Description

Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string

S is dominated by

T if

S is a substring of

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

3
10
you
better
worse
richer
poorer
sickness
health
death
faithfulness
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
5
abc
cde
abcde
abcde
bcde
3
aaaaa
aaaab
aaaac

Sample Output

youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
abcde
No
#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<string.h>
#include<map> 
#include<math.h>
#include<set>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
int nex[100010];
int* buildNext(string &P)
{
	unsigned j = 0, m =P.size();
	int t = nex[0] = -1;
	while (j < m)
	{
		if (t < 0 || P[j] == P[t])
			nex[++j] = ++t;
		else
			t = nex[t];
	}
	return nex;
}
// 在 T 中寻找 P
int Match(string &P, string &T)
{
	int *next = buildNext(P);
	int i = 0, n = T.size();
	int j = 0, m = P.size();
	while (i < n && j < m)
	{
		if (j < 0 || T[i] == P[j])
			++i, ++j;
		else
			j = next[j];
		if (j >= m)
			return 1;
	}
	return 0;
}
int kmpCount(string &P, string &T)//出现次数 
{
	int i, n = T.size();
	int j, m = P.size();
	int *next = buildNext(P);
	int _count = 0;
	i = j = 0;
	while (i < n && j < m)
	{
		if (0 > j || T[i] == P[j])
			i++, j++;
		else
			j = next[j];
		if (j == m)
		{
			_count++;
			j = 0;
		}
	}
	delete[] next;
	return _count;
}
int main()
{
    int n,i,k,j,flag,max=0,biao;
    string s[100005],S,T;
    scanf("%d",&n);
    while(n--)
    {
    	max=0;
        scanf("%d",&k);
        for(i=1;i<=k;i++)
        {
          cin>>s[i];
		  if(s[i].size()>max)
		  {
		  	max=s[i].size();//找出最长的
		  	biao=i;
		  }	
		}
		  S=s[biao];
            flag=0;
            for(j=1;j<=k;j++)
            {
            	
            	T=s[j];
                if(!Match(T,S))//匹配
                {
                    flag=1;
                    break;
                }
            }
            if(flag==0)
            {
                cout<<S<<endl; 
            }
            else
            printf("No\n");
    }
}