A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is
the tail substring of “asdf” and the head substring of the “sdfg” . However, the result comes as “asdfghjk”, when you have to add “asdf” and “ghjk” and guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

Input

For each case, there are two strings (the chars selected just form ‘a’ to ‘z’) for you, and each length of theirs won’t exceed 10^5 and won’t be empty.

Output

Print the ultimate string by the book.

Sample Input

asdf sdfg
asdf ghjk

Sample Output

asdfg
asdfghjk

Author

Wang Ye

Source

2008杭电集训队选拔赛——热身赛

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| Note

这题我是用扩展kmp做的, 做两次就行

/*************************************************************************
    > File Name: hdu1867.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年02月02日 星期一 17时03分34秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 200110;

char S[N], T[N];
int next[N];
int extend[N];
char A1[N], A2[N];

void EXTEND_KMP (char S[], char T[])
{
	int lens = strlen (S);
	int lent = strlen (T);
	next[0] = lent;
	int i, j, p, L;
	j = 0;
	while (j + 1 < lent && T[j] == T[j + 1]) // 先求出next[1]
	{
		++j;
	}
	next[1] = j;
	int a = 1;
	for (i = 2; i < lent; ++i)
	{
		p = next[a] + a - 1;
		L = next[i - a];
		if (i + L < p + 1)
		{
			next[i] = L;
		}
		else
		{
			j = max (0, p - i + 1);
			while (i + j < lent && T[i + j] == T[j])
			{
				++j;
			}
			next[i] = j;
			a = i;
		}
	}
	j = 0;
	while (j < lens && S[j] == T[j])
	{
		++j;
	}
	extend[0] = j;
	a = 0;
	for (i = 1; i < lens; ++i)
	{
		p = extend[a] + a - 1;
		L = next[i - a];
		if (L + i < p + 1)
		{
			extend[i] = L;
		}
		else
		{
			j = max(0, p - i + 1);
			while (i + j < lens && j < lent && S[i + j] == T[j])
			{
				++j;
			}
			extend[i] = j;
			a = i;
		}
	}
}

int main ()
{
	while (~scanf("%s%s", S, T))
	{
		EXTEND_KMP (S, T);
		int lens = strlen (S);
		int lent = strlen (T);
		int s = -1;
		strcpy (A1, S);
		for (int i = 0; i < lens; ++i)
		{
			if (extend[i] + i == lens)
			{
				s = i;
				break;
			}
		}
		if (s == -1)
		{
			strcat (A1, T);
		}
		else
		{
			int cnt = lens;
			for (int i = lens - s; i < lent; ++i)
			{
				A1[cnt++] = T[i];
			}
			A1[cnt] = '\0';
		}
		memset (next, 0, sizeof(next));
		memset (extend, 0, sizeof(extend));
		strcpy (A2, T);
		EXTEND_KMP (T, S);
		s = -1;
		for (int i = 0; i < lent; ++i)
		{
			if (extend[i] + i == lent)
			{
				s = i;
				break;
			}
		}
		if (s == -1)
		{
			strcat (A2, S);
		}
		else
		{
			int cnt = lent;
			for (int i = lent - s; i < lens; ++i)
			{
				A2[cnt++] = S[i];
			}
			A2[cnt] = '\0';
		}
		int len1 = strlen (A1);
		int len2 = strlen (A2);
		if (len1 < len2)
		{
			printf("%s\n", A1);
		}
		else if (len1 > len2)
		{
			printf("%s\n", A2);
		}
		else
		{
			if (strcmp (A1, A2) < 0)
			{
				printf("%s\n", A1);
			}
			else
			{
				printf("%s\n", A2);
			}
		}
	}
	return 0;
}