hdu 1867 A + B for you again(KMP灵活应用)

vocaloid01

于 2019-02-28 18:14:50 发布

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分类专栏： Kmp

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本文介绍了一种基于KMP算法的字符串处理方法，用于解决两个字符串A和B的重合部分省略问题，通过实现A+B和B+A的最短字符串输出。利用KMP算法进行字符串匹配，找到可以压缩的最大字符数，最终选择可以压缩更多字符的字符串组合，并在长度相等时选择字典序最小的输出。

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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

题意：两个字符串A，B进行重合部分省略一边（例如 ABC+BC=ABC , ABC+BCD=ABCD，ABCD+BC=ABCDBC）的A+B和B+A得到两个串，然后输出其中较短的（长度相同时输出字典序最小的）。

题解：两个串互相KMP得到A+B和B+A各可以压缩多少字母，然后选择可以压缩的多的，相同的话比较字典序就行了。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 100005;

int _next[MAXN];
char s[MAXN],S[MAXN];

void getNext(char board[],int L){
    int k = -1;
    int j = 0;
    _next[0] = -1; 
    while(j<L){
        if(k == -1 || board[j] == board[k]){
            if(board[j+1] == board[k+1]){ 
                _next[++j] = _next[++k];
            }
            else _next[++j] = ++k;
        }
        else {
            k = _next[k];
        }
    }
}

int Find(char board[],char str[],int L1,int L2){//灵活运用点主要在这个函数
    getNext(board,L1);
    int temp1 = max(L2-L1,0),temp2 = 0;
    while(temp1<L2){
        if(temp2 == -1 || str[temp1] == board[temp2]){
            temp1++;temp2++;
        }
        else{
            temp2 = _next[temp2];
        }
    } 
    
    return temp2;
}

int main(){
	
	while(scanf("%s %s",s,S) != EOF){
		int a,b;
		a = Find(s,S,strlen(s),strlen(S));
		b = Find(S,s,strlen(S),strlen(s));
		//printf("#####%d %d  %d %d\n",a.ads,a.num,b.ads,b.num);
		if(a == b){
			if(strcmp(s,S) <= 0)printf("%s%s\n",s,S+b);
			else printf("%s%s\n",S,s+a);
		}
		else if(a < b)printf("%s%s\n",s,S+b);
		else  printf("%s%s\n",S,s+a);
	}
	
	return 0;
}