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

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string>
 4 #include<algorithm>
 5 #include<vector>
 6 using namespace std;
 7 int _,n,Next[110];
 8 vector<string> t;
 9
10 void prekmp(string s) {
11     int len=s.size();
12     int i,j;
13     j=Next[0]=-1;
14     i=0;
15     while(i<len) {
16         while(j!=-1&&s[i]!=s[j]) j=Next[j];
17         Next[++i]=++j;
18     }
19 }
20
21 int kmp(string t,string p) {
22     int lent=t.size(),lenp=p.size();
23     int i=0,j=0,ans=-1,res;
24     res=-1;
25     while(i<lent) {
26         while(j!=-1&&t[i]!=p[j]) j=Next[j];
27         ++i;++j;
28         res=max(res,j);
29     }
30     ans=max(ans,res);
31     reverse(t.begin(),t.end());
32     res=-1;
33     i=0,j=0;
34     while(i<lent) {
35         while(j!=-1&&t[i]!=p[j]) j=Next[j];
36         ++i;++j;
37         res=max(res,j);
38     }
39     ans=max(ans,res);
40     return ans;
41 }
42
43 int main() {
44    // freopen("in","r",stdin);
45     for(scanf("%d",&_);_;_--) {
46         scanf("%d",&n);
47         string s;
48         for(int i=0;i<n;i++) {
49             cin>>s;
50             t.push_back(s);
51         }
52         string str=t[0],tempstr;
53         int len=str.size(),maxx=-1;
54         for(int i=0;i<len;i++) {
55             tempstr=str.substr(i,len-i);
56             int ans=0x3f3f3f;
57             prekmp(tempstr);
58             for(int j=1;j<t.size();j++) {
59                 ans=min(kmp(t[j],tempstr),ans);
60             }
61             maxx=max(maxx,ans);
62         }
63         printf("%d\n",maxx);
64         t.clear();
65     }
66 }

原文地址：https://www.cnblogs.com/ACMerszl/p/10299885.html