HDU：2594-Simpsons’ Hidden Talents

Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

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

---

解题心得：

1. 题目又说了一大堆的废话，其实就是问你字符串s1的前缀和字符串s2的后缀能匹配的最大的长度是多少。
2. 和裸的KMP差不多了。

---

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<string>
using namespace std;
const int maxn = 5e4+100;
int Next[maxn];
string s1,s2;

void cal_next() {
    int k = -1;
    Next[0] = -1;
    for (int i = 1; i < s1.size(); i++){
        while(k>-1 && s1[i] != s1[k+1])
            k = Next[k];
        if(s1[i] == s1[k+1])
            k++;
        Next[i] = k;
    }
}

void init() {
    memset(Next,-1, sizeof(Next));
    cal_next();//先处理s1，用s1去匹配s2
}

void get_ans(){
    bool flag = false;
    int k = -1;
    for (int i = 0; i <s2.size() ; ++i) {
        while(s1[k+1] != s2[i] && k>-1)
            k = Next[k];
        if(s1[k+1] == s2[i])
            k++;
        if(i+1 == s2.size()){//已经匹配到了s2的最后一位
            if(k == -1)//和最后一位都不匹配，匹配长度为0，flag为标记
                break;
            flag = true;
            for(int j=k;j>=0;j--)//将匹配的长度为k的字符输出来
                printf("%c",s2[s2.size()-j-1]);
            printf(" %d\n",k+1);
            return ;
        }
    }
    if(!flag)
        printf("0\n");
}

int main(){
    while(cin>>s1>>s2){
        init();
        get_ans();
    }
    return 0;
}