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hdu 2594 kmp水題 求s1的前綴和s2的后綴重復(fù)度的最大值

2013-06-05 11:16?1199人閱讀?評(píng)論(0)?收藏?舉報(bào) ?分類： KMP（8）?

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Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1588????Accepted Submission(s): 587


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 飛機(jī)票：http://acm.hdu.edu.cn/showproblem.php?pid=2594 題意：? 輸入 s1 s2? 問s1的后綴 以及s2的前綴中 相同的字符最多有多少個(gè)? 是什么 思路：? kmp [cpp]?view plaincopy

1. #include<stdio.h>??
2. #include<string.h>??
3. char?s1[51111],s2[51111];??
4. int?next[51111];??
5. int?d1,d2;??
6. void?getnext()??
7. {??
8. ????int?i=1,j=0;??
9. ????next[1]=0;??
10. ????while(i<=d1)??
11. ????{??
12. ????????if(j==0||s1[i]==s1[j])?{i++;j++;next[i]=j;}??
13. ????????else?j=next[j];??
14. ????}??
15. }??
16. int?kmp()??
17. {??
18. ????int?i,j;??
19. ????i=1;j=1;??
20. ????while(i<=d2)??
21. ????{??
22. ????????if(j==0||s2[i]==s1[j])?{i++;j++;}??
23. ????????//不用擔(dān)心這里j會(huì)超出其最大長度d2?因?yàn)楫?dāng)j==d2+1的時(shí)候?就會(huì)進(jìn)入j=next[j]返回前面去了??
24. ????????//因?yàn)閟1[d2+1]是沒有值的不會(huì)等于對應(yīng)位置的s2中的值??會(huì)返回到前面去??
25. ????????else?j=next[j];??
26. ????}??
27. ????return?j-1;??
28. }??
29. int?main()??
30. {??
31. ????int?k;??
32. ????while(scanf("%s",s1+1)!=EOF)??
33. ????{??
34. ????????scanf("%s",s2+1);??
35. ????????d1=strlen(s1+1);??
36. ????????d2=strlen(s2+1);//s2做主串??
37. ????????getnext();??
38. ????????k=kmp();??
39. ????????if(k==0)?printf("0\n");??
40. ????????else??
41. ????????printf("%s?%d\n",s2+1+d2-k,k);??
42. ????}??
43. ????return?0;??
44. } ?
45. 下面是另一種做法

46. 題意：

    給你兩個(gè)串a(chǎn),b，求既是a的前綴又是b的后綴的最長子串的長度。

    分析：

    很自然的想到把兩個(gè)串連接起來，根據(jù)KMP的性質(zhì)求即可

    #include <map>
    #include <set>
    #include <list>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <cstdio>
    #include <vector>
    #include <string>
    #include <cctype>
    #include <complex>
    #include <cassert>
    #include <utility>
    #include <cstring>
    #include <cstdlib>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    typedef pair<int,int> PII;
    typedef long long ll;
    #define lson l,m,rt<<1
    #define pi acos(-1.0)
    #define rson m+1,r,rt<<11
    #define All 1,N,1
    #define N 50010
    #define read freopen("in.txt", "r", stdin)
    const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
    const int INF= 0x7ffffff;
    const int mod =  1000000007;
    char a[N*2],b[N];
    int f[N*2];
    void getnext(int n){int i=0,j=-1;f[0]=-1;while(i<n){if(j==-1||a[i]==a[j]){i++;j++;f[i]=j;}elsej=f[j];}
    }
    void solve(){int len1=strlen(a);int len2=strlen(b);strcat(a,b);int tmp=len1+len2;getnext(tmp);//必須是兩串的子串while(f[tmp]>len1)tmp=f[tmp];while(f[tmp]>len2)tmp=f[tmp];if(f[tmp]==0)printf("0\n");else{for(int i=0;i<f[tmp];++i)printf("%c",a[i]);printf(" %d\n",f[tmp]);}
    }
    int main()
    {while(~scanf("%s%s",a,b)){solve();}
    return 0;
    }

    
    

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