HDU 2594 Simpsons’ Hidden Talents

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5061    Accepted Submission(s): 1845



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
 
题意是求最长主串的前缀与模式串后缀匹配。
这种,只要连接后,便可利用KMP的功能,来实现找到其前缀后缀的最大匹配。
注意中间需要一个不会出现的字符进行隔开,以防出现自我匹配的情况。

代码如下;
/*************************************************************************
	> File Name: Simpsons_Hidden_Talents.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Wed 02 Dec 2015 05:14:12 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

void preKMP(char x[], int m, int kmpnext[]){
    int i, j;
    i = 0;
    j = kmpnext[0] = -1;
    while(i < m){
        while(j != -1 && x[i] != x[j])
            j = kmpnext[j];
        if(x[++ i] == x[++ j])
            kmpnext[i] = kmpnext[j];
        else 
            kmpnext[i] = j;
    }
}

int nexti[100010];

int main(void){
    char str1[100010];
    char str2[50010];
    while(~(scanf("%s%s", str1, str2))){
        int temp1 = strlen(str1);
        str1[temp1] = '&';
        str1[temp1 + 1] = '\0';
        strcat(str1, str2);
        int temp = strlen(str1);
        preKMP(str1, temp, nexti);
        if(nexti[temp]){
            for(int i  = 0; i < nexti[temp]; i ++){
                printf("%c", str1[i]);
            }
            cout << " ";
            cout << nexti[temp] << endl;
        }
        else
            cout << nexti[temp] << endl;
    }
}


posted @ 2015-12-03 00:14  ChiLuManXi  阅读(167)  评论(0编辑  收藏  举报