HDU 1867 A + B for you again （kmp）

A + B for you again

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8929    Accepted Submission(s): 2163

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

思路：分别查看一个字符串的后缀是不是另一个字符串的前缀，然后取最大长度的

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define mod 1000000007
#define mem(a) memset(a,0,sizeof(a))

using namespace std;

const int maxn = 100000 + 5 , inf = 0x3f3f3f3f;

int Next[maxn];
char str1[maxn],str2[maxn];
void cal_next(char *a){
    memset(Next,-1,sizeof(Next));
    int len = strlen(a);
    Next[0] = -1;
    int k = -1 ;
    for(int p = 1 ; p < len ; p ++ ){
        while(k>-1&&a[k+1]!=a[p]) k = Next[k];
        if(a[k+1]==a[p])
            Next[p]=++k;
    }
}
int kmp(char *s,char *a){
    cal_next(a);
    int slen = strlen(s);
    int alen = strlen(a);
    int k = -1;
    int i = 0;
    for(; i < slen ; i++ ){
        while(k>-1&&s[i]!=a[k+1]) k = Next[k];
        if(a[k+1]==s[i]) k++;
    }
    if(i==slen){
//        cout<<i<<" "<<k<<endl;
        return k;
    }
    return -1;
}
void solve(){
    int x = kmp(str1,str2);
    int y = kmp(str2,str1);
    if(x>y){
        printf("%s",str1);
        printf("%s\n",str2+x+1);
    }
    else if(x<y){
        printf("%s",str2);
        printf("%s\n",str1+y+1);
    }
    else{
        if(strcmp(str1,str2)<0){
            printf("%s",str1);
            printf("%s\n",str2+x+1);
        }
        else{
            printf("%s",str2);
            printf("%s\n",str1+y+1);
        }
    }
}
int main(){
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    while(scanf("%s %s",str1,str2)!=EOF) solve();
    return 0 ;
}