hdu1238 Substrings 扩展KMP

Posted 2020-09-05 奚政

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.

 

找出最长的一个子串，所有串中都出现过它或它的逆序

扩展KMP水题

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 #include<string>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 const int maxn=505;
 9 int nxt[105][maxn],ext[105][maxn];
10 char s[105][maxn];
11 int len[105],ans[maxn];
12 
13 void EKMP(char s[],char t[],int lens,int lent,int c){
14     int i,j,p,l,k;
15     nxt[c][0]=lent;j=0;
16     while(j+1<lent&&t[j]==t[j+1])j++;
17     nxt[c][1]=j;
18     k=1;
19     for(i=2;i<lent;i++){
20         p=nxt[c][k]+k-1;
21         l=nxt[c][i-k];
22         if(i+l<p+1)nxt[c][i]=l;
23         else{
24             j=max(0,p-i+1);
25             while(i+j<lent&&t[i+j]==t[j])j++;
26             nxt[c][i]=j;
27             k=i;
28         }
29     }
30 
31     j=0;
32     while(j<lens&&j<lent&&s[j]==t[j])j++;
33     ext[c][0]=j;k=0;
34     for(i=1;i<lens;i++){
35         p=ext[c][k]+k-1;
36         l=nxt[c][i-k];
37         if(l+i<p+1)ext[c][i]=l;
38         else{
39             j=max(0,p-i+1);
40             while(i+j<lens&&j<lent&&s[i+j]==t[j])j++;
41             ext[c][i]=j;
42             k=i;
43         }
44     }
45 }
46 
47 int main(){
48     int T;
49     scanf("%d",&T);
50     while(T--){
51         memset(ans,0x3f,sizeof(ans));
52         int n;
53         scanf("%d",&n);
54         for(int i=1;i<=n;++i)scanf("%s",s[i]);
55         for(int i=1;i<=n;++i)len[i]=strlen(s[i]);
56         s[1][len[1]]=\'#\';
57         for(int i=1;i<=len[1];++i)s[1][len[1]+i]=s[1][len[1]-i];
58         s[1][len[1]+len[1]+1]=0;
59         len[1]=strlen(s[1]);
60         int maxx=0;
61         for(int i=0;i<len[1];++i){
62             for(int j=2;j<=n;++j){
63                 int cnt=0;
64                 EKMP(s[j],s[1]+i,len[j],len[1]-i,j);
65                 for(int k=0;k<len[j];++k){
66                     if(ext[j][k]>cnt)cnt=ext[j][k];
67                 }
68                 if(cnt<ans[i])ans[i]=cnt;
69             }
70             if(ans[i]>maxx)maxx=ans[i];
71         }
72         printf("%d\\n",maxx);
73     }
74     return 0;
75 }

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