HDU 5510 Bazinga 字符串HASH (2015ACM/ICPC亚洲区沈阳站)

【题目链接】：click here~~

【题目大意】：

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 39    Accepted Submission(s): 15

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
   
   
   
   
   
   
   
   


   
   
   
   

求前i个字符串不是该i字符串的子串的最大的i值

【思路】暴力KMP，暴力字符串HASH

此处留坑代填~~

先放挫挫代码：

/*
* Problem: HDU No.5510
* Running time: 399MS
* Complier: G++
* Author: javaherongwei
* Create Time: 17:30 2015/10/31 星期六
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef unsigned long long LLU;
const LLU base=1e9+7;
const int N=505;
const int M=2020;
char s[N][M];
LLU f[N][M], fac[M];
vector<int> val;
int n, len[N];
LLU get(int i, int l, int r)
{
    return f[i][r]-f[i][l-1]*fac[r-l+1];
}
bool check(int x, int y)
{
    if(len[y]>len[x]) return false;
    for(int i = len[y]; i <= len[x]; i++)
    {
        if(f[y][len[y]] == get(x,i-len[y]+1,i)) return true;
    }
    return false;
}
int solve()
{
    for(int i = 0; i < n; i++)
    {
        len[i] = strlen(s[i]+1);
        f[i][0] = 0;
        for(int j = 1; j <= len[i]; j++)
        {
            f[i][j] = f[i][j-1]*base+s[i][j]-'a'+1;
        }
    }
    int ans = -1;
    val.clear();
    val.push_back(0);
    for(int i = 1; i < n; i++)
    {
        while(val.size())
        {
            if(check(i, val[val.size()-1]))
            {
                val.pop_back();
            }
            else  break;
        }
        val.push_back(i);
        if(val.size()>1) ans = i+1;
    }
    return ans;
}
void init()
{
    fac[0] = 1;
    for(int i = 1; i <= 2002; i++) fac[i] = fac[i-1]*base;
}
int main()
{
    //freopen("1.txt","r",stdin);
    init();
    int t, tot = 0;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%s", s[i]+1);
        printf("Case #%d: %d\n", ++tot, solve());
    }
    return 0;
}