HDU 5510 Bazinga

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 101    Accepted Submission(s): 41

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）

 解题：KMP暴力搞，如果A是B的子串，B是C的子串，那么A一定是C的子串。如果B不是C的子串，A有可能是C的子串。

这说明了什么？这说明如果A是B的子串，那么A就没用了，这就是一个除去冗余的数据的过程，接下来就是暴力了。由于题目的要求是它之前的必须是它的子串，所以，只要比较相邻的串即可除去冗余的串

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2010;
int fail[maxn];
void getFail(char str[]) {
    for(int i = 0,j = fail[0] = -1; str[i]; ++i) {
        while(j != -1 && str[i] != str[j]) j = fail[j];
        fail[i + 1] = ++j;
    }
}
bool match(char sa[],char sb[]) {
    getFail(sa);
    for(int i = 0,j = 0; sb[i]; ++i) {
        while(j != -1 && sb[i] != sa[j]) j = fail[j];
        if(!sa[++j]) return true;
    }
    return false;
}
char S[501][maxn];
bool invalid[501];
int main() {
    int kase,n,cs = 1;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%d",&n);
        memset(invalid,false,sizeof invalid);
        for(int i = 0; i < n; ++i) {
            scanf("%s",S[i]);
            if(i && match(S[i-1],S[i]))
                invalid[i-1] = true;
        }
        int ret = -1;
        for(int i = n-1; i >= 0 && ret == -1; --i) {
            for(int j = 0; j < i && ret == -1; ++j) {
                if(invalid[j]) continue;
                if(!match(S[j],S[i])) ret = i + 1;
            }
        }
        printf("Case #%d: %d
",cs++,ret);
    }
    return 0;
}