Find the Shortest Common Superstring
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1081 Accepted Submission(s): 276
Problem Description
The shortest common superstring of 2 strings S1 and S2 is a string S with the minimum number of characters which contains both S1 and S2 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.
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 S1 and the second line contains the string S2. Both
of these strings contain at least 1 and at most 1.000.000 characters.
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
题目是求包含两个串的字符串的最小长度
/*
思路:求最小的字符串能包含s1,s2两个字符串,则最多需要s1+s2构成的串
如果s1,s2能重叠的话(s1包含s2,s2包含s1或s1,s2部分重叠),即
aaaabb
bbccc
这样需要的串就更短了,所以可以用s1去匹配s2求得s1前缀和s2后缀重叠数
用s2去匹配s1求得s2前缀和s1后缀重叠数
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;
const int MAX=1000000+10;
char s1[MAX],s2[MAX];
int next[MAX];
void get_next(char *a,int len){
int i=-1,j=0;
next[0]=-1;
while(j<len){
if(i == -1 || a[i] == a[j]){
if(a[++i] == a[++j])next[j]=next[i];
else next[j]=i;
}else i=next[i];
}
return;
}
int KMP(char *a,char *b,int lena,int lenb){
int i=0,j=0;
while(i<lena && j<lenb){
if(i == -1 || a[i] == b[j])++i,++j;
else i=next[i];
}
return i;
}
int main(){
int n;
cin>>n;
while(n--){
scanf("%s%s",s1,s2);
int lena=strlen(s1),lenb=strlen(s2);
get_next(s1,lena);
int a=KMP(s1,s2,lena,lenb);
get_next(s2,lenb);
int b=KMP(s2,s1,lenb,lena);
if(a == lena)cout<<lenb<<endl;
else if(b == lenb)cout<<lena<<endl;
else cout<<lena+lenb-max(a,b)<<endl;
}
return 0;
}