KMP的next数组性质运用

hdu2594

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2525    Accepted Submission(s): 960

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

分析：

next[j]=k：下标为j的前k个字符和开头的前k个字符完全一样

对于字符串abcuijkabc的长度为10，即next[10]=3；

所以这道题可以先把两个字符串合并，用next数组的性质可以更高效的求出；

注意：

aaaa和aaa答案是aaaaaa 6不符合题意

在合并的时候在两个字符串之间加上一个特殊字符如“#”

即新字符串为aaaa#aaa，此时的答案为 aaa 3

程序：

#include"stdio.h"
#include"string.h"
#define M 100009
int next[M];
void getNext(char *b)
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while(b[i]!='')
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
char ch[M],ch1[M];
int main()
{
    while(scanf("%s%s",ch,ch1)!=EOF)
    {
        char str[M];
        strcat(ch,"0");
        strcat(ch,ch1);
        int m=strlen(ch);
        getNext(ch);
        strncpy(str,ch,next[m]);
        str[next[m]]='';//注意
        if(next[m])
        printf("%s %d
",str,next[m]);
        else
        printf("0
");
    }
}