HDU5510(Bazinga）(KMP)

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6280    Accepted Submission(s): 1927

Problem Description

Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.

For  n  given strings  S1,S2,⋯,Sn , labelled from  1  to  n , you should find the largest  i (1≤i≤n)  such that there exists an integer  j (1≤j<i)  and  Sj  is not a substring of  Si .

A substring of a string  Si  is another string that occurs  in  Si . For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".

 

Input

The first line contains an integer  t (1≤t≤50)  which is the number of test cases.
For each test case, the first line is the positive integer  n (1≤n≤500)  and in the following  n  lines list are the strings  S1,S2,⋯,Sn .
All strings are given in lower-case letters and strings are no longer than  2000  letters.

 

Output

For each test case, output the largest label you get. If it does not exist, output  −1 .

 

Sample Input

  

   4
5
ab
abc
zabc
abcd
zabcd
4
you
lovinyou
aboutlovinyou
allaboutlovinyou
5
de
def
abcd
abcde
abcdef
3
a
ba
ccc
  

 

Sample Output

  

   Case #1: 4
Case #2: -1
Case #3: 4
Case #4: 3
  

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

字符串匹配，找到最大的字符串下标使得存在字符串不是它的字串

每两个相邻的字符串用比较，则可以获得相邻的两个字符串是不是互为字串，最后再用二重循环搜索一遍找到下标最大的字符串即可

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char str[505][2005];
int ne[5005];
bool visit[505];

void getnex(char p[])
{
    ne[0]=0;
    int n=strlen(p);
    for(int i=1,k=0;i<n;i++)
    {
        while(k>0&&p[k]!=p[i])
        {
            k=ne[k-1];
        }
        if(p[k]==p[i])
        {
            k++;
        }
        ne[i]=k;
    }
}
int KMP(char p[],char t[])
{
        getnex(p);
        int n=strlen(p);
        int ans=0;
        int len=strlen(t);
        for(int i=0,q=0;i<len;i++)
        {
            while(q>0&&p[q]!=t[i])
                q=ne[q-1];
            if(p[q]==t[i])
            {
                q++;
            }
            if(q==n)
            {
                ans++;
                q=0;
            }
        }
        return ans;
}

int main()
{
    int t;
    int ca=1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        memset(visit,true,sizeof(visit));
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%s",str[i]);

        for(int i=1;i<n;i++)
        {
            if(KMP(str[i-1],str[i])>0)
            {
                visit[i-1]=0;
            }

        }
        printf("Case #%d: ",ca++);
        int flag=0;
        for(int i=n-1;i>=0;i--)
        {
            for(int j=0;j<i;j++)
            {
                if(visit[j])
                {
                    if(KMP(str[j],str[i])==0)
                       {
                           printf("%d\n",i+1);
                            flag=1;
                           break;
                       }
                }
            }
            if(flag==1)
                break;
        }
        if(flag==0)
            printf("-1\n");
    }
    return 0;
}