hdu1841Find the Shortest Common Superstring

Find the Shortest Common Superstring

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2062    Accepted Submission(s): 570

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

2albabacauresitamures

 

Sample Output

78
让一个串的前缀与令一个串的后缀匹配 其实与之前模式串与主串的匹配没有什么不同 还是模式串在一直后移 只是超过长度时返回已匹配的长度 
#include <iostream>#include<cstdio>#include<cstring>using namespace std;char a[1000005],b[1000005];int n[1000005];int len1,len2;void kmp(char *t,int len){    n[0]=-1;    int k=-1;    int j=0;    while(j<=len)    {        if(k==-1||t[j]==t[k])        {            k++;            j++;            n[j]=k;        }        else k=n[k];    }}int solve(char *c, char *d){    int i=0,j=0;    int tmp1=strlen(c),tmp2=strlen(d);    kmp(d,tmp2);    while(i<tmp1&&j<tmp2)    //j<tmp2 是d是c的主串那种情况    {        if(j==-1||c[i]==d[j])i++,j++;        else j=n[j];    }    return j;}int main(){    int t;    cin>>t;    while(t--)    {        scanf(" %s %s",a,b);        len1=strlen(a);        len2=strlen(b);        int ans1=solve(a,b);        int ans2=solve(b,a);        cout<<(len1+len2-max(ans1,ans2))<<endl;    }    return 0;}