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

題意：每組數據給出兩個字符串 s1、s2，判斷是否存在?s1 的前綴是 s2 的后綴的情況，若沒有則輸出 0，若有則輸出這個前綴/后綴與其長度

思路：KMP 的應用

本質上就是找前一個字符串的前綴和后一個字符串的后綴相同的串，使用 KMP 進行匹配時，注意當模式串 s1 到達結尾時，需要更新指針位置，返回值也需要返回前綴位置

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #define PI acos(-1.0) #define E 1e-9 #define INF 0x3f3f3f3f #define LL long long const int MOD=20091226; const int N=50001; const int dx[]= {-1,1,0,0}; const int dy[]= {0,0,-1,1}; using namespace std;int Next[N]; char str1[N],str2[N]; void getNext(char p[]){Next[0]=-1;int len=strlen(p);int j=0;int k=-1;while(j<len){if(k==-1||p[j]==p[k]) {k++;j++;Next[j]=k;}else{k=Next[k];}} } int KMP(char t[],char p[]){int tLen=strlen(t);int pLen=strlen(p);int i=0;int j=0;while(i<tLen&&j<pLen){if(j==-1||t[i]==p[j]){i++;j++;if(j==pLen&&i!=tLen){//當模式串到達結尾時，回到指定位置j=Next[j];}}else{j=Next[j];}}return j;//返回前綴的位置 }int main(){while(scanf("%s",str1)!=EOF){scanf("%s",str2);getNext(str1);int res=KMP(str2,str1);if(res!=0){str1[res]='\0';//截取前綴printf("%s %d\n",str1,res);}elseprintf("0\n");}return 0; }

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總結

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