kuangbin专题十六 KMP&&扩展KMP HDU1238 Substrings

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

InputThe first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.
OutputThere should be one line per test case containing the length of the largest string found.
Sample Input

2
3
ABCD
BCDFF
BRCD
2
rose
orchid

Sample Output

2
2



这道题和之前一道题类似，都是多个字符串匹配问题，只是这道题多了倒序的匹配。直接reverse就可以了
遍历第一个串的所有后缀，然后匹配每个字符串的正反序。得到公共匹配的最小。然后枚举所有后缀取最大

#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
int _,n,Next[110];
vector<string> t;

void prekmp(string s) {
    int len=s.size();
    int i,j;
    j=Next[0]=-1;
    i=0;
    while(i<len) {
        while(j!=-1&&s[i]!=s[j]) j=Next[j];
        Next[++i]=++j;
    }
}

int kmp(string t,string p) {
    int lent=t.size(),lenp=p.size();
    int i=0,j=0,ans=-1,res;
    res=-1;
    while(i<lent) {
        while(j!=-1&&t[i]!=p[j]) j=Next[j];
        ++i;++j;
        res=max(res,j);
    }
    ans=max(ans,res);
    reverse(t.begin(),t.end());
    res=-1;
    i=0,j=0;
    while(i<lent) {
        while(j!=-1&&t[i]!=p[j]) j=Next[j];
        ++i;++j;
        res=max(res,j);
    }
    ans=max(ans,res);
    return ans;
}

int main() {
   // freopen("in","r",stdin);
    for(scanf("%d",&_);_;_--) {
        scanf("%d",&n);
        string s;
        for(int i=0;i<n;i++) {
            cin>>s;
            t.push_back(s);
        }
        string str=t[0],tempstr;
        int len=str.size(),maxx=-1;
        for(int i=0;i<len;i++) {
            tempstr=str.substr(i,len-i);
            int ans=0x3f3f3f;
            prekmp(tempstr);
            for(int j=1;j<t.size();j++) {
                ans=min(kmp(t[j],tempstr),ans);
            }
            maxx=max(maxx,ans);
        }
        printf("%d\n",maxx);
        t.clear();
    }
}