这个题和kmp算法的共同点，也就是可以用kmp解的原因，在于当前缀所在串（kmp中的模式串）字符pj≠后缀所在串(kmp中文本串)字符tj时，应使前缀串(kmp中模式串)尽量往右移动最大位移，而暴力算法则是每次移动位移为1。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>using namespace std;#define maxn 50005char t[maxn], p[maxn];
int next[maxn];void getNext(int *next, char *p, int size){int j=0, k=next[0]=-1;while(j<size){if(k==-1 || p[j]==p[k]) next[++j]=++k;else k=next[k];}
}int kmp(char *t, int lt, char *p, int lp){int i=0, j=0;while(i<lt && j<lp){if(j==-1 || t[i]==p[j]) i++, j++;else j=next[j];}return j;           //没找到j=0，否则j=length of common substring
}int main(){while(scanf(" %s %s", p, t)==2){int lt = strlen(t),  lp = strlen(p);char *text = lt > lp ? lt-lp+t : t;lt = min(lt, lp);getNext(next, p, lp);int k = kmp(text, lt, p, lp);if(!k) printf("0\n");else{p[k]=0;printf("%s %d\n", p, k);}}return 0;
}

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2650 Accepted Submission(s): 1004

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

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转载于:https://www.cnblogs.com/ramanujan/p/3737739.html

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