HDU 5510：Bazinga（暴力KMP）

http://acm.hdu.edu.cn/showproblem.php?pid=5510

Bazinga

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

题意：给出n个串，求出最大的i使得有一个j（1<=j<i）不是i的子串。

思路：比赛的时候由于没剩下多少时间，写的太暴力了TLE，后面自己写了一个，发现别人写的也很暴力，还有用strstr过的，速度普遍比用KMP的快。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 2010
#define N 510
int nxt[M];
char s[N][M];

void make_next(char *s)
{
    memset(nxt, 0, sizeof(nxt));
    int j = -1, i = 0;
    nxt[0] = -1;
    int len = strlen(s);
    while(i < len) {
        if(j == -1 || s[i] == s[j]) {
            i++; j++;
            nxt[i] = j;
        } else {
            j = nxt[j];
        }
    }
}

int kmp(char *s, char *str)
{
    make_next(s);
    int l = strlen(s), len = strlen(str);
    int i = 0, j = 0;
    while(i < len && j < l) {
        if(j == -1 || str[i] == s[j]) {
            i++; j++;
            if(j >= l) return true;
        } else {
            j = nxt[j];
        }
    }
    return false;
}

int main()
{
    int t;
    int cas = 0;
    scanf("%d", &t);
    while(t--) {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%s", s[i]);

        int tmp = -100;
        for(int i = n; i > 1; i--){
            if(!kmp(s[i-1], s[i])) {
                 tmp = i;
                 break;
            }
        }
//        printf("TMP : %d
", tmp);
        int ans = tmp;
        for(int i = tmp - 1; i > 0; i--) {
            while(!kmp(s[i], s[ans+1]) && ans < n) {
                ans++;
            }
        }

        printf("Case #%d: ", ++cas);
        if(tmp != -100)
            printf("%d
", ans);
        else
            printf("-1
");
    }
    return 0;
}