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最新推荐文章于 2019-01-24 16:04:00 发布

Jade1998

最新推荐文章于 2019-01-24 16:04:00 发布

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分类专栏：字符串匹配

本文链接：https://blog.youkuaiyun.com/qq_37605573/article/details/78152071

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本文介绍了一个字符串匹配问题，目标是找出最长的序列使得该序列中不存在任何一个之前的字符串作为其子串。文章提供了详细的解决思路和完整的代码实现。

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Problem Description

Ladies and gentlemen, please sit up straight.
Don't tilt your head. I'm serious.

For

n given strings

S 1 ,S 2 ,⋯,S n , 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

S j is not a substring of

S i .

A substring of a string

S i is another string that occurs in

S i . 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

S 1 ,S 2 ,⋯,S n .
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：求第i个字符串不是 1—i 个字符串（存在一个匹配不成就行），求 i 的最大值

首先定义一个vis[i][j]数组用来存放第i个字符串与第j个字符串匹配情况，mapp[i]数组表示第i个字符串是之前字符串的子串

i从1开始 j从i-1开始匹配，如果第i个字符串与第j个字符串匹配成功，就更新vis数组，寻找可能不匹配的字符串在此匹配；否则，mapp[i]标记，退出当前循环。

然后从大到小判断第i个字符串与之前的是否匹配

若i==0，则输出-1即可，否则输出i+1

代码如下：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

const int N=500+5;
const int NMAX=2000+5;
int vis[N][N],mapp[N];
int next_s[NMAX];
char s[N][NMAX];
void get_next(int next_s[],char s[],int len)
{
    int i=0,j;
    j=next_s[0]=-1;
    while(i<len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++,j++;
            next_s[i]=j;
        }
        else
            j=next_s[j];
    }
}
int kmp(char s[],char t[],int ls,int lt)
{
    int i=0,j=0;
    int ans=0;
    while(i<ls)
    {
        if(j==-1 || s[i]==t[j])
            i++,j++;
        else
            j=next_s[j];
        if(j>=lt){
            ans++;
            break;
        }
    }
    return ans;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    int count=1;
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        memset(mapp,1,sizeof(mapp));
        scanf("%d",&n);
        scanf("%s",s[0]);
        for(int i=1;i<n;i++)
        {
            scanf("%s",s[i]);
            int len_si=strlen(s[i]);
            for(int j=i-1;j>=0;j--)
            {
                if(j==i-1)
                {
                    memset(next_s,0,sizeof(next_s));
                    int len_sj=strlen(s[j]);
                    get_next(next_s,s[j],len_sj);
                    int ans=kmp(s[i],s[j],len_si,len_sj);
                    if(ans!=0)//匹配成功
                    {
                        for(int k=0;k<j;k++)
                            vis[i][k]=vis[i][k]|vis[j][k];
                        vis[i][j]=1;
                    }
                    else
                    {
                        mapp[i]=0;
                        break;
                    }

                }
                else
                {
                    if(!vis[i][j])
                    {
                        int len_sj=strlen(s[j]);
                        memset(next_s,0,sizeof(next_s));
                        get_next(next_s,s[j],len_sj);
                        int ans=kmp(s[i],s[j],len_si,len_sj);
                        if(ans!=0)//匹配成功
                        {
                            for(int k=0;k<j;k++)
                                vis[i][k]=vis[i][k]|vis[j][k];
                            vis[i][j]=1;
                        }
                        else
                        {
                            mapp[i]=0;
                            break;
                        }
                    }
                }
            }
        }
        int i;
        for(i=n-1;i>=1;i--)
            if(mapp[i]==0)
                break;
        if(i==0)
            printf("Case #%d: -1\n",count++);
        else
            printf("Case #%d: %d\n",count++,i+1);
    }
    return 0;
}