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[550][2200];
/*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,0,sizeof(f));
    int len=strlen(p);  /* only cal once, or TLE */
    for(int i=1;i<len;i++)
    {
        int j=f[i];
        while(j&&p[i]!=p[j]) j=f[j];
        f[i+1]=(p[i]==p[j]? j+1:0);
    }
}
int kmp(char *str,char *p)    /*O(len1+len2)*/
{
    getf(p);
    cnt=0;
    int len1=strlen(str),len2=strlen(p);    /* only cal once, or TLE*/
    for(int i=0,j=0;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!=0) break;
    }
    return cnt;
}

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

    int t,ca=1;scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%s",Str[i]);

        int flag[maxn]={0};
        for(int i=1;i<n;i++)
            if(kmp(Str[i+1],Str[i])!=0) flag[i]=1;

        for(int i=n;i>=1;i--){
            for(int j=1;j<i;j++){
                if(flag[j]) continue;
                kmp(Str[i],Str[j]);
                if(cnt==0){
                    printf("Case #%d: %d\n",ca++,i);goto s;
                }
            }
        }

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

    return 0;
}