HDU 5510 Bazinga KMP

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4518    Accepted Submission(s): 1442

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）

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<double, double> PII;

const int MX = 2000 + 5;
const int MX2 = 500 + 5;

/*
void GetNext() {
   Next[0] = 0;
   for(int i = 1; i < n; i++) {
       int j = Next[i - 1];
       while(j && S[i] != S[j]) j = Next[j - 1];
       Next[i] = S[i] == S[j] ? j + 1 : 0;
   }
}
*/
int Next[MX], n;
int KMP(char *A, char *B) {
    int m = strlen(A), n = strlen(B);
    Next[0] = 0;
    for(int i = 1; i < n; i++) {
        int k = Next[i - 1];
        while(B[i] != B[k] && k) k = Next[k - 1];
        Next[i] = B[i] == B[k] ? k + 1 : 0;
    }
    int ans = 0, j = 0;
    for(int i = 0; i < m; i++) {
        while(A[i] != B[j] && j) j = Next[j - 1];
        if(A[i] == B[j]) j++;
        if(j == n) ans++;
    }
    return ans;
}

char ch[MX2][MX];
bool vis[MX2];

int main() {
    //FIN
    int T;
    int cnt = 1;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        memset(vis, 0, sizeof(vis));
        for(int i = 1; i <= n; i++) scanf("%s", ch[i]);
        for(int i = 2; i <= n ;i++)
            if(KMP(ch[i], ch[i-1])) vis[i-1] = 1;

        int flag = 0;
        for(int i = n; i >= 1; i--) {
            for(int j = 1; j < i; j++) {
                if(!vis[j])
                if(KMP(ch[i], ch[j]) == 0) {
                    flag = 1;
                    printf("Case #%d: %d
", cnt++, i);
                    break;
                }
            }
            if(flag) break;
        }
        if(!flag) printf("Case #%d: -1
", cnt++);
    }
    return 0;
}