【hdu 2594】Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8010 Accepted Submission(s): 2837

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

【题解】

很巧妙的题;
把s1当成匹配串;
s2当成被匹配串;
在s2中匹配s1;
然后做KMP;
可知到了s1的最后一个字符的时候匹配到s2中的第几个字符;
则这些字符的个数就是题目所需要的长度;
(做KMP的时候，每次往移动都是最优的.匹配出来的肯定也是最长的了;)
做KMP的时候始终保持匹配串的前缀即s1[1..j]是和s2[i-j+1..i]相同的;这里i是被匹配串扫描到的位置;
如果i==length(s2)了;
那么s2[i-j+1..i]就是s2的后缀了即s2[len2-j+1..len2]
而在做KMP的时候会让j最大吧?

【完整代码】

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 5e4+10;

string s1,s2;
int l1,l2;
int f[MAXN];

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    while (cin >> s1 >> s2)
    {
        l1 = s1.size(),l2 = s2.size();
        s1 = ' '+s1,s2 = ' '+s2;
        //s1 zicuan s2 zhucuan
        f[1] = f[2] = 1;
        for (int i = 2;i <= l1;i++)
        {
            int t = f[i];
            while (t>1 && s1[i]!=s1[t]) t = f[t];
            f[i+1] = s1[i]==s1[t]?t+1:1;
        }
        //kmp
        int j = 1;
        for (int i = 1;i <= l2;i++)
        {
            while (j>1 && s2[i]!=s1[j]) j = f[j];
            if (s2[i]==s1[j])
                j++;
            if (j>l1)
            {
                if (i==l2)
                    break;
                j = f[j];
            }
        }
        if (j==1)
            puts("0");
        else
        {
            for (int i = 1;i <= j-1;i++)
                putchar(s1[i]);
            putchar(' ');
            printf("%d
",j-1);
        }
    }
    return 0;
}