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
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.
The lengths of s1 and s2 will be at most 50000.