Hdu-1867 A + B for you againA + B for you again

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A + B for you again

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6027 Accepted Submission(s): 1493

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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分析：两次KMP

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#define max(a,b) a > b ? a : b
using namespace std;
char a[100005],b[100005],s1[200005],s2[200005];
int Next[100005];
void got(char a[],char b[],char c[])
{
	int n = strlen(a),m = strlen(b);
	memset(Next,0,sizeof(Next));
	int now = Next[0] = -1;
	for(int i = 1;i < m;i++)
	{
		while(now >= 0 && b[now+1] != b[i]) now = Next[now];
		if(b[now+1] == b[i]) now++;
		Next[i] = now;
	}
	now = -1;
	for(int i = max(0,n-m);i < n;i++)
	{
		while(now >= 0 && b[now+1] != a[i]) now = Next[now];
		if(b[now+1] == a[i]) now++;
	}
	memset(c,0,sizeof(c));  
	for(int i = 0;i < n;i++) c[i] = a[i];
	int l = n - 1;
	for(int i = now + 1;i <= m;i++) c[++l] = b[i];
}
int main()
{
	cin.sync_with_stdio(false);
	while(cin>>a>>b)
	{
		got(a,b,s1);
		got(b,a,s2);
		if(strlen(s1) > strlen(s2) || (strlen(s1) == strlen(s2) && strcmp(s1,s2) > 0)) cout<<s2<<endl;
		else cout<<s1<<endl;
	}
}