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  

 

Recommend lcy

 

題意：

給定字串$s1$$s2$，在$s1$中找一個字首和$s2$的字尾匹配的長度最長。

思路：

$next$陣列的定義。

所以把$s1$$s2$拼起來求$next$就可以了。需要考慮一下越過他們的邊界的情況。

 

 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 }