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.
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.
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
Sample Output
题目要求2个字符串的shortest common superstring，最小公共父串：使得str1和str2都包含在所求的superstring中，且该superstring的长度最小。设str1的长度为m，str2的长度为n，如果按照逐个位置匹配求公共部分的方法，时间复杂度为O(m*n)。题目中的n∈[1,1000000]，显然会TLE，必须降低时间复杂度，只能用KMP模式匹配算法来做了，时间复杂度O(m+n)。
用KMP算法来求解时，分3种情况：
----str1与str2能完全匹配，长度为max{len1,len2},如banana，ban；
----str1与str2不能匹配，但是能找到公共部分，如字符串alba，bacau，公共部分ba长度为2；
----str1与str2不能匹配，且没有公共部分，如asdf，ghjk。
1 #include <iostream>
2
3 #define max(a,b) (a>b?a:b)
4 #define match(a,b) ((a)==(b))
5 const int MAXN = 1000001;
6 int fail[MAXN];
7 char str[2][MAXN];
8
9 int kmp_pat_match(char *str,int ls,int &i,char *pat,int lp,int &j){
10 fail[0]=-1,i=0,j;
11 for(j=1;j<lp;j++){
12 for(i=fail[j-1];i>=0 && !match(pat[i+1],pat[j]);i=fail[i]);
13 fail[j]=match(pat[i+1],pat[j])?i+1:-1;
14 }
15 for(i=j=0;i<ls&&j<lp;i++)
16 if(match(str[i],pat[j])) j++;
17 else if(j)
18 j=fail[j-1]+1,i--;
19 return j==lp?(i-lp):-1;
20 }
21 int main(){
22 int i,j,t,u,v,len1,len2,pos;
23 scanf("%d",&t),getchar();
24 while(t--){
25 gets(str[0]),gets(str[1]);
26 len1=strlen(str[0]),len2=strlen(str[1]);
27 u=0,pos=kmp_pat_match(str[0],len1,i,str[1],len2,j);
28 if(pos!=-1) u=len2;
29 else if(i==len1 && j>0 && match(str[0][len1-1],str[1][j-1])) u=j;
30 v=0,pos=kmp_pat_match(str[1],len2,i,str[0],len1,j);
31 if(pos!=-1) v=len1;
32 else if(i==len2 && j>0 && match(str[1][len2-1],str[0][j-1])) v=j;
33 printf("%d\n",len1+len2-max(u,v));
34 }
35 return 0;
36 }