HDU 1867 A + B for you again 数据结构+KMP简单应用

A + B for you again 数据结构+KMP简单应用

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

SubmitStatus

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 sdfgasdf ghjk 

 

Sample Output

asdfgasdfghjk 解题思路：看懂这题的题意，看到了复杂度之后，首先确定不能用普通的暴力算法来做主要的目标还是求出他们结合的位数，因为这两个串先后顺序不确定，那么就要两个都求出来，然后找一个最优的怎么求出来结合的位数？假如结合的是A,B串，那我就把B当作模式串，A当作待匹配的串那么我就让B去匹配A，当A匹配到最后一个的时候，然后获得此时的模式串下标，该下标就是结合的位数。
#include<cstdio>#include<cstring>#include<iostream>using namespace std;const int maxn = 100005 ;char text[maxn] ;char str[maxn] ;int f[maxn] ;char ans1[2*maxn] ;char ans2[2*maxn] ;int find(char *T,char *P){    int ans = 0 ;    int n = strlen(T);    int m = strlen(P);    int j = 0 ;    for(int i=0;i<n;i++){        while(j&&P[j]!=T[i])j=f[j] ;        if(P[j]==T[i])j++ ;        if(i==n-1){            return j;        }    }    return ans ;}void getfail(char *P){    int n = strlen(P) ;    f[0] = 0 ;    f[1] = 0 ;    for(int i=1;i<n;i++){        int j = f[i] ;        while(j&&P[i]!=P[j])j=f[j] ;        f[i+1] = (P[i]==P[j]?j+1:0) ;    }    return ;}int main(){    while(~scanf("%s%s",text,str)){        memset(f,0,sizeof(f));        getfail(str) ;        int k = find(text,str) ;        //printf("k = %d\n",k);        strcpy(ans1,text) ;        strcat(ans1,str+k) ;        //printf("%s\n",ans1);        memset(f,0,sizeof(f));        getfail(text) ;        k = find(str,text) ;        strcpy(ans2,str) ;        strcat(ans2,text+k) ;        //printf("%s\n",ans2);        if(strlen(ans1)==strlen(ans2)){            strcmp(ans1,ans2)<0?puts(ans1):puts(ans2) ;        }else{            strlen(ans1)>strlen(ans2)?puts(ans2):puts(ans1) ;        }    }    return 0;}