HDU 5510 Bazinga（KMP）

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4572    Accepted Submission(s): 1459

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

 

题意
找到一个最大的i，使得i前面的字符串至少有一个不是它的子串

 

分析：如果直接用KMP进行匹配操作就会超时，这里需要剪枝的是如果前面的字符串已经是另一个字符串的子串了
     就不需要再把它拿来进行匹配。

     一开始用的是vector来存需要匹配的字符串，但是这样的话，在存入和取出的时候用时也很高

     后面就换 vis来记录不用匹配的下标，就过了

代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;

const int N = 1000002;
int Next[N];
char S[N], T[N];
char str[510][2100];
int slen, tlen;//注意每次一定要计算长度
int len[510];
int vis[510];
vector<int>V;
void getNext(char *T,int tlen)
{
    int j, k;
    j = 0; k = -1; Next[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
            Next[++j] = ++k;
        else
            k = Next[k];

}
/*
返回模式串T在主串S中首次出现的位置
返回的位置是从0开始的。
*/
int KMP_Index(char *S,int slen,char *T,int tlen)
{
    int i = 0, j = 0;
    getNext(T,tlen);

    while(i < slen && j < tlen)
    {
        if(j == -1 || S[i] == T[j])
        {
            i++; j++;
        }
        else
            j = Next[j];
    }
    if(j == tlen)
        return i - tlen;
    else
        return -1;
}
int main()
{

    int TT,Case=0;;
    int i, cc,n,sum,maxx;
    scanf("%d",&TT);
    while(TT--)
    {
        memset(vis,0,sizeof(vis));
        maxx=-1;
        Case++;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%s",str[i]);
             len[i]=strlen(str[i]);
        }
        for(int i=1;i<n;i++)
        {
          sum=0;
          for(int j=0;j<i;j++)
          {
             if(!vis[j]){
             if(KMP_Index(str[i],len[i],str[j],len[j])>=0)
              vis[j]=1;
            else
            {
                maxx=i;
                break;
            }
          }
          }
        }
        if(maxx==-1)  printf("Case #%d: %d
",Case,maxx);
        else printf("Case #%d: %d
",Case,maxx+1);
    }
    return 0;
}