题目1841 Find the Shortest Common Superstring（KMP）

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/48048555

本文探讨了两个字符串的最短公共超字符串问题，并提供了一种通过KMP算法寻找最优解的方法。该算法能有效确定两个字符串组合成最短长度的新字符串，使其同时包含原始字符串作为子序列。

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Find the Shortest Common Superstring

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1629 Accepted Submission(s): 437

Problem Description

The shortest common superstring of 2 strings S ₁ and S ₂ is a string S with the minimum number of characters which contains both S ₁ and S ₂ as a sequence of consecutive characters. For instance, the shortest common superstring of “alba” and “bacau” is “albacau”.
Given two strings composed of lowercase English characters, find the length of their shortest common superstring.

Input

The first line of input contains an integer number T, representing the number of test cases to follow. Each test case consists of 2 lines. The first of these lines contains the string S ₁ and the second line contains the string S ₂. Both of these strings contain at least 1 and at most 1.000.000 characters.

Output

For each of the T test cases, in the order given in the input, print one line containing the length of the shortest common superstring.

Sample Input

  
  
   
   2
alba
bacau
resita
mures

Sample Output

7
8

Author

Mugurel Ionut Andreica

Source

Politehnica University of Bucharest Local Team Contest 2007

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lcy | We have carefully selected several similar problems for you: 1801 1823 1427 1839 1582

应该说很水的，就有考虑一下，一个串包含另一个串的情况就好了

ac代码

#include<stdio.h>
#include<string.h>
int next[1000100];
char ans[2000100],s1[1000100],s2[1000100];
void getnext(char *s,int len)
{
	next[0]=-1;
	int i,j;
	i=0;
	j=-1;
	while(i<=len)
	{
		if(j==-1||s[i]==s[j])
		{
			i++;
			j++;
			next[i]=j;
		}
		else
			j=next[j];
	}
}
int kmp(char *a,char *b,int len1,int len2)
{	
//	memset(next,0,sizeof(next));
	getnext(b,len2);
	int i,j;
	i=j=0;
	while(i<len1)
	{
		if(j==-1||a[i]==b[j])
		{
			i++;
			j++;
		}
		else
			j=next[j];
		if(j==len2)
			return len2;
	}
	if(i==len1)
		return j;
	return 0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s",s1,s2);
		int len1=strlen(s1);
		int len2=strlen(s2);
		int x=kmp(s1,s2,len1,len2);
		int y=kmp(s2,s1,len2,len1);
		int ans;
		if(x>y)
		{
			ans=len1+len2-x;
		}
		else
		{
			ans=len1+len2-y;
		}
		printf("%d\n",ans);
	}
}