Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15015    Accepted Submission(s): 5151

Problem Description

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Input

Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

Output

Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input

clinton
homer
riemann
marjorie

Sample Output

0
rie 3

Source

HDU 2010-05 Programming Contest

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lcy

题意：

给定字符串$s1$$s2$，在$s1$中找一个前缀和$s2$的后缀匹配的长度最长。

思路：

$next$数组的定义。

所以把$s1$$s2$拼起来求$next$就可以了。需要考虑一下越过他们的边界的情况。

 1 #include<iostream>
 2 //#include<bits/stdc++.h>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstdlib>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<vector>
10 #include<set>
11 #include<climits>
12 #include<map>
13 using namespace std;
14 typedef long long LL;
15 typedef unsigned long long ull;
16 #define pi 3.1415926535
17 #define inf 0x3f3f3f3f
18
19 const int maxn = 50005;
20 char s1[maxn * 2], s2[maxn];
21 int nxt[maxn * 2];
22
23 void getnxt(char *s)
24 {
25     int len = strlen(s);
26     nxt[0] = -1;
27     int k = -1;
28     int j = 0;
29     while(j < len){
30         if(k == -1 || s[j] == s[k]){
31             ++k;++j;
32             if(s[j] != s[k]){
33                 nxt[j] = k;
34             }
35             else{
36                 nxt[j] = nxt[k];
37             }
38         }
39         else{
40             k = nxt[k];
41         }
42     }
43 }
44
45 bool kmp(char *s, char *t)
46 {
47     getnxt(s);
48     int slen = strlen(s), tlen = strlen(t);
49     int i = 0, j = 0;
50     while(i < slen && j < tlen){
51         if(j == -1 || s[i] == t[j]){
52             j++;
53             i++;
54         }
55         else{
56             j = nxt[j];
57         }
58     }
59     if(j == tlen){
60         return true;
61     }
62     else{
63         return false;
64     }
65 }
66
67 int main()
68 {
69     while(scanf("%s", s1) != EOF){
70         //getchar();
71         scanf("%s", s2);
72         //cout<<s1<<endl<<s2<<endl;
73         int len1 = strlen(s1), len2 = strlen(s2);
74         strcat(s1, s2);
75         //cout<<s1<<endl;
76         getnxt(s1);
77         //cout<<nxt[len1 + len2]<<endl;
78         if(nxt[len1 + len2] > min(len1, len2)){
79             for(int i = 0; i < min(len1, len2); i++){
80                 printf("%c", s1[i]);
81             }
82             printf(" %d\n", min(len1, len2));
83         }
84         else{
85             int ans = nxt[len1 + len2];
86             if(ans){
87                 for(int i = 0; i < ans; i++){
88                     printf("%c", s1[i]);
89                 }
90                 printf(" ");
91             }
92
93             printf("%d\n", ans);
94         }
95     }
96     return 0;
97 }

hdu2594 Simpsons' Hidden Talents【next数组应用】

原文地址：https://www.cnblogs.com/wyboooo/p/10034300.html