Bazinga

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 593    Accepted Submission(s): 241

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，s[i] (1≤i≤n),找到最大的i，是得存在j(1≤j<i)使得，s[j]不是 s[i]的子串。

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

char s[550][2200];
int Next[550][2200];

int Judge(int w, int o)
{
    int i = 0 , j = 0;
    int l1, l2;
    l1 = strlen(s[w]);
    l2 = strlen(s[o]);

    while(i < l1 && j < l2)
    {
        if(s[w][i] == s[o][j] || i == -1)
        {
            i++;
            j++;
        }
        else
            i = Next[w][i];

        if(i == l1)
        {
            return true;
        }
    }
    return false;
}

void getNext(int q)
{
    int j, k;
    int i = strlen(s[q]);
    j = 0;
    k = -1;
    Next[q][0] = -1;

    while(j < i)
    {
        if(k == -1 || s[q][j] == s[q][k])
        {
            j++;
            k++;
            Next[q][j] = k;
        }
        else
            k = Next[q][k];
    }
}

int main()
{
    int t, n, index, l = 1;
    scanf("%d", &t);

    while(t--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%s", s[i]);
            getNext(i);
        }
        index = -1;
        int q;
        for(int i = n-2; i >= 0; i--)
        {
            if(!Judge(i, i+1))  找到一个字符串不满足前后是子串关系后，从后向前找最大的不满足子串的串
            {
                q = i;
                for(int j = n-1; j > q; j--)
                {
                    if(!Judge(q, j))
                        index = index > j ? index : j;
                 }
            }
        }
        if(index == -1)
            printf("Case #%d: %d\n", l++, index);
        else
            printf("Case #%d: %d\n", l++, index+1);
    }
    return 0;
}

　　strstr阔以考虑……脉动回来

    学长递归写法……

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n, num;
char s[550][2200];

void slove(int k)
{
    if(k == 0 || num == n-1)
        return ;
    if(strstr(s[k], s[k-1]))
        slove(k-1);   // 向下寻找
    else
    {
        for(int i = n-1; i >= num; i--)   // 找最大的 下标
        {
            if(strstr(s[i], s[k-1]) == 0)
            {
                num = i;
                break;     // 
            }
        }
        slove(k-1);   // 继续向下寻找看是否有 存在 k 使 num 最大  
    }
}

int main()
{
    int t, l = 1;
    scanf("%d", &t);

    while(t--)
    {
        scanf("%d", &n);
        num = -1;
        for(int i = 0; i < n; i++)
            scanf("%s", s[i]);
        slove(n-1);
        printf("Case #%d: %d\n", l++, num == -1 ? num : num+1);
    }
    return 0;
}