A + B for you again
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4633 Accepted Submission(s): 1192
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.
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
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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: zchao1995@gmail.com
> 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;
}