题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1867

A + B for you again

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

kmp。。。

注意: 字符串A+B也可是B+A反正输出相加之后最短的那个，若相加之后长度相等输出字典序最小的那个。。

 #include<algorithm>

 #include<iostream>

 #include<cstdlib>

 #include<cstring>

 #include<cstdio>

 #include<vector>

 #include<map>

 using std::cin;

 using std::cout;

 using std::endl;

 using std::find;

 using std::sort;

 using std::map;

 using std::pair;

 using std::vector;

 using std::multimap;

 #define pb(e) push_back(e)

 #define sz(c) (int)(c).size()

 #define mp(a, b) make_pair(a, b)

 #define all(c) (c).begin(), (c).end()

 #define iter(c) decltype((c).begin())

 #define cls(arr,val) memset(arr,val,sizeof(arr))

 #define cpresent(c, e) (find(all(c), (e)) != (c).end())

 #define rep(i, n) for (int i = 0; i < (int)(n); i++)

 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)

 const int N = ;

 typedef unsigned long long ull;

 int next[N];

 char str1[N], str2[N];

 struct KMP {

     int i, j, n, m;

     inline void get_next(char *src) {

         m = strlen(src);

         for (i = , j = next[] = ; i < m; i++) {

             while (j >  && src[i] != src[j]) j = next[j - ];

             if (src[i] == src[j]) j++;

             next[i] = j;

         }

     }

     inline int kmp_match(char *text, char *pat) {

         n = strlen(text);

         for (i = j = ; i < n; i++) {

             while (j >  && text[i] != pat[j]) j = next[j - ];

             if (text[i] == pat[j]) j++;

         }

         return j;

     }

 }go;

 int main() {

 #ifdef LOCAL

     freopen("in.txt", "r", stdin);

     freopen("out.txt", "w+", stdout);

 #endif

     int p, q;

     while (~scanf("%s %s", str1, str2)) {

         go.get_next(str2);

         p = go.kmp_match(str1, str2);

         go.get_next(str1);

         q = go.kmp_match(str2, str1);

         if (p > q || (p == q && - == strcmp(str1, str2))) {

             printf("%s%s\n", str1, str2 + p);

         } else {

             printf("%s%s\n", str2, str1 + q);

         }

     }

     return ;

 }

巴特西

hdu 1867 A + B for you again