hdu 2594 Simpsons’ Hidden Talents

给出字符串S1和S2求最长字符串，使之即是S1的前缀又是S2的后缀。

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5203    Accepted Submission(s): 1883

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

clintonhomerriemannmarjorie

 

Sample Output

0rie 3

 

Source

HDU 2010-05 Programming Contest

 

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解法一：用一次KMP即可。KMP匹配时指向缝衣针内的指针一定表示指针前面已经匹配。KMP算法已经蕴含了最长。

/**========================================== *   This is a solution for ACM/ICPC problem * *   @source：hdu 2594 Simpsons’ Hidden Talents *   @type:  data_structure KMP *   @author: wust_ysk *   @blog:  http://blog.csdn.net/yskyskyer123 *   @email: [e-mail] *===========================================*/#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;const int maxn=50000    ;char N[maxn+10];char M[maxn+10];int nex[maxn+10];int len_M;void getnex(){    nex[0]=nex[1]=0;    for(int i=1;i<len_M;i++)    {        int p=nex[i];        while(p&&M[p]!=M[i]) p=nex[p];        nex[i+1]=M[p]==M[i]?p+1:0;    }}void KMP(){    int len_N=strlen(N);    int p=0;    for(int i=0;i<len_N;i++)    {        while(p&&N[i]!=M[p])  p=nex[p];        if(i==len_N-1)  break;        if(N[i]==M[p])  p++;        if(p==len_M)  p=nex[p];    }    int len=N[len_N-1]==M[p]?p+1:0;//  int len=N[len_N-1]==M[p]?p+1:p;一样的，如果=了p，p必定为0    if(!len)  {puts("0");return;}    for(int i=len_N-1-len+1;i<len_N;i++)    {        putchar(N[i]);    }    printf(" %d",len);    putchar('\n');}int main(){    while(~scanf("%s%s",M,N))    {        len_M=strlen(M);        getnex();        KMP();    }   return 0;}

解法二：利用next[]，将S2放在S1后面求解。

/**========================================== *   This is a solution for ACM/ICPC problem * *   @source：hdu 2594 Simpsons’ Hidden Talents *   @type:  data_structure KMP *   @author: wust_ysk *   @blog:  http://blog.csdn.net/yskyskyer123 *   @email: [e-mail] *===========================================*/#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;const int maxn=50000    ;char N[maxn+10];char M[2*maxn+10];int nex[2*maxn+10];int len_M;int len_N;int len;void getnex(){    nex[0]=nex[1]=0;    for(int i=1;i<len;i++)    {        int p=nex[i];        while(p&&M[p]!=M[i]) p=nex[p];        nex[i+1]=M[p]==M[i]?p+1:0;    }    int ans=min(len_M,nex[len]);//     ans=min(len_N,ans);//这两步很重要    if(!ans)  {puts("0");return;}    for(int i=0;i<ans;i++)    {        putchar(M[i]);    }    putchar(' ');    printf("%d\n",ans);}int main(){    while(~scanf("%s%s",M,N))    {        len_M=strlen(M);        len_N=strlen(N);        len =len_M+len_N;        for(int i=0;i<len_N;i++)        {            M[len_M+i]=N[i];        }        M[len]='\0';        getnex();    }   return 0;}