Simpsons’ Hidden Talents

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2798    Accepted Submission(s): 1055

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

这题碰到了一些莫名其妙的问题，next数组本是记录模式串本身的匹配信息，我想着能不能用在两个不同的串上，然后就试验了一下，结果能想到的测试数据都能通过，但是就是WA，至今不知道错在哪里。

题意：给定两个串，求第二个串的后缀跟第一个串的前缀能匹配的最大长度并输出这个匹配串。

题解：在网上看到一种思路是将第一个串之后添加一个特殊字符，再把第二个串粘在第一个串后面，然后对这个新串求next数组，最终next[len]即为所求。

#include <stdio.h>
#include <string.h>
#define maxn 50002

char str1[maxn << 1], str2[maxn];
int ans, next[maxn << 1], len2;

void getNext()
{
	int i = 0, j = -1;
	next[0] = -1;
	while(str1[i]){
		if(j == -1 || str1[i] == str1[j]){
			++i; ++j;
			next[i] = j;
		}else j = next[j];
	}
	str1[ans = next[i]] = '\0';
}

int main()
{
	//freopen("stdin.txt", "r", stdin);
	while(scanf("%s%s", str1, str2) == 2){
		strcat(str1, " ");
		strcat(str1, str2); getNext();
		ans ? printf("%s %d\n", str1, ans) : printf("0\n");
	}
}

放一个典型的错误代码，能想到的测试数据都通过了，但是提交就WA。不知道为什么。

#include <stdio.h>
#define maxn 50002

char str1[maxn], str2[maxn] = {' '};
int ans, next[maxn];

void getNext()
{
    int i = 0, j = -1;
    next[0] = -1;
    while(str2[i]){
        if(j == -1 || str2[i] == str1[j]){
            ++i; ++j;
            next[i] = j;
        }else j = next[j];
    }
    str1[ans = next[i]] = '\0';
}

int main()
{
    //freopen("stdin.txt", "r", stdin);
    while(scanf("%s%s", str1, str2 + 1) == 2){
        getNext();
        ans ? printf("%s %d\n", str1, ans) : printf("0\n");
    }
}

另外一个不知道错哪里的代码：

#include <stdio.h>
#define maxn 50002

char str1[maxn], str2[maxn] = {' ', ' '};
int ans, next[maxn];

void getNext()
{
    int i = 0, j = -1;
    next[0] = -1;
    while(str2[i]){
        if(j == -1 || str2[i] == str1[j]){
            ++i; ++j;
            next[i] = j;
        }else j = next[j];
    }
    str1[ans = next[i]] = '\0';
}

int main()
{
    //freopen("stdin.txt", "r", stdin);
    while(scanf("%s%s", str1, str2 + 2) == 2){
        getNext();
        ans ? printf("%s %d\n", str1, ans) : printf("0\n");
    }
}

HDU2594 Simpsons’ Hidden Talents 【KMP】