hdu 2594 Simpsons’ Hidden Talents

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原文链接：http://www.cnblogs.com/ACFLOOD/p/4346180.html

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博客给出题目链接，分析判断两串中是否存在能作为s1前缀且s2后缀的子串问题。可将两串合并，利用next数组，但合并后需判断所求子串是否大于s1或s2的长度。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2594

分析：判断是否在两串中存在子串，能作为s1的前缀 && s2的后缀。可以考虑将两串合并，然后直接利用next数组。合并后可能会出现的状况是，所求子串大于s1 || s2的长度，所以还要进行一次判断。

/*Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3522    Accepted Submission(s): 1313


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
*/
#include <cstdio>
#include <cstring>
const int maxn = 50000 + 10;
int n, m, next[2*maxn];
char s1[maxn], s2[maxn], p[2*maxn];
void getNext()
{
    next[0] = -1;
    int j = 0, k = -1;
    while(j < n+m){ //遍历合并之后的p串 
        if(k == -1 || p[k] == p[j]) next[++j] = ++k;
        else k = next[k];
    }
}
int main()
{
    while(~scanf("%s", s1)){
        scanf("%s", s2);
        n = strlen(s1); m = strlen(s2);
        memcpy(p, s1, sizeof(s1));
        strcat(p, s2);
        getNext();
        if(!next[n+m]) printf("0\n"); //不存在直接输出0 
        else{
            int cnt = next[n+m];
            s1[cnt] = '\0';
            if(n > m) {//比较两串的长度 
                if(cnt > m) printf("%s %d\n", s2, m); //超出最小串长直接输出  
                else printf("%s %d\n", s1, cnt);
            }
            else {
                if(cnt > n) printf("%s %d\n", s1, n);
                else printf("%s %d\n", s1, cnt);
            }
        }
    }
    return 0;
}