HDU 5510 Bazinga (2015沈阳现场赛，子串判断)

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 214    Accepted Submission(s): 89

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）

 

 

【题意】：

n个字符串，找出最大的i，使得存在j 满足j<i 且Sj不是Si的字串。

 

【解题思路】：

看着是个水题本想直接上，看到一片TLE吓得上了个KMP的板。。。

交了一发还是TLE，已经不想做了。。

在某blog的指点下，先判断相邻两个串是否存在字串关系，若Si是Sj的字串，则i和j只需要kmp()一次即可~

减少kmp()次数后800ms过了，，感觉还是不够优美

搜了一下发现有各种做法，Hash、dp......

有空再改，挖坑待填~~

 

(代码写得丑==)

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<string>
 #include<algorithm>
 #define eps 1e-8
 #define zero(x)(((x)>0?(x):-(x))<eps)
 #define PI acos(-1.0)
 #define LL long long
 #define maxn 2200
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 char Str[][];
 /*KMP--字符串匹配--单一模版串*/
 //char str[maxn],p[maxn];/*p is template string*/
 int f[maxn],cnt;//when failed,get to f[i];
 void getf(char *p)
 {
     memset(f,,sizeof(f));
     int len=strlen(p);  /* only cal once, or TLE */
     for(int i=;i<len;i++)
     {
         int j=f[i];
         while(j&&p[i]!=p[j]) j=f[j];
         f[i+]=(p[i]==p[j]? j+:);
     }
 }
 int kmp(char *str,char *p)    /*O(len1+len2)*/
 {
     getf(p);
     cnt=;
     int len1=strlen(str),len2=strlen(p);    /* only cal once, or TLE*/
     for(int i=,j=;i<len1;i++)
     {
         while(j&&str[i]!=p[j]) j=f[j];
         if(str[i]==p[j]) j++;
         if(j==len2) cnt++;/*Match success*/
         if(cnt!=) break;
     }
     return cnt;
 }

 int main(int argc, char const *argv[])
 {
     //IN;

     int t,ca=;scanf("%d",&t);
     while(t--)
     {
         int n;scanf("%d",&n);
         for(int i=;i<=n;i++) scanf("%s",Str[i]);

         int flag[maxn]={};
         for(int i=;i<n;i++)
             if(kmp(Str[i+],Str[i])!=) flag[i]=;

         for(int i=n;i>=;i--){
             for(int j=;j<i;j++){
                 if(flag[j]) continue;
                 kmp(Str[i],Str[j]);
                 if(cnt==){
                     printf("Case #%d: %d\n",ca++,i);goto s;
                 }
             }
         }

         printf("Case #%d: -1\n",ca++);
         s:continue;
     }

     return ;
 }