BNUOJ34990--Justice String （exkmp求最长公共前缀）

Justice String

Given two strings A and B, your task is to find a substring of A called justice string, which has the same length as B, and only has at most two characters different from B.

 

Input

The first line of the input contains a single integer T, which is the number of test cases.

For each test case, the first line is string A, and the second is string B.

Both string A and B contain lowercase English letters from a to z only. And the length of these two strings is between 1 and 100000, inclusive. 

 

 

Output

For each case, first output the case number as "Case #x: ", and x is the case number. Then output a number indicating the start position of substring C in A, position is counted from 0. If there is no such substring C, output -1.

And if there are multiple solutions, output the smallest one. 

 

Sample Input

3
aaabcd
abee
aaaaaa
aaaaa
aaaaaa
aabbb

Sample Output

Case #1: 2
Case #2: 0
Case #3: -1

题意：两个字符串， 求B在在A串出现的第一个位置， 可以允许最多两个字符不同。
做法： 依次枚举位置i， 然后求A[i, ...lena - 1]与B的最长公共前缀， 再求 A[i, i+lenb-1]的B的最长公共后缀。。 然后对于中间那一部分， 用多项式hash直接判断是否相等。

 #include <bits/stdc++.h>
 using namespace std;
 const int seed = 1e9+;
 const int MAXN = 1e5+;
 typedef unsigned long long uLL;
 uLL _hash[][MAXN];
 string s1, s2;
 void Hash(string s, int d){
     int len = s.size();
     memset(_hash[d], , sizeof (_hash[d]));
     for (int i = ; i < len; i++){
         _hash[d][i] = (i ? _hash[d][i-] : ) * seed + s[i] - '';
     }
 }
 void pre_kmp(string &s, int m, int next[]){
     next[] = m;
     int j = ;
     while (j +  < m && s[j] == s[j+]){
         j++;
     }
     next[] = j;
     int k = ;
     for (int i = ; i < m; i++){
         int p = next[k] + k - ;
         int L = next[i-k];
         if (i + L < p + ){
             next[i] = L;
         }else{
             j = max(, p-i+);
             while (i+j < m && s[i+j] == s[j]){
                 j++;
             }
             next[i] = j;
             k = i;
         }
     }
 }
 void exkmp(string &str1, int len1, string &str2, int len2, int next[], int extend[]){
     pre_kmp(str1, len1, next);
     int j = ;
     while (j < len2 && j < len1 && str1[j] == str2[j]){
         j++;
     }
     extend[] = j;
     int k = ;
     for (int i = ; i < len2; i++){
         int p = extend[k] + k - ;
         int L = next[i-k];
         if (i + L < p + ){
             extend[i] = L;
         }else{
             j = max(, p-i+);

             while (i+j < len2 && j < len1 && str2[i+j] == str1[j]){
                 j++;
             }
             extend[i] = j;
             k = i;
         }
     }
 }
 int ex1[MAXN], ex2[MAXN], next[MAXN];
 uLL bas[MAXN];
 void pre_solve(){
     bas[] = ;
     for (int i = ; i < MAXN; i++){
         bas[i] = bas[i-] * seed;
     }
 }
 int main(){

     //freopen("in.txt", "r", stdin);
     pre_solve();
     int T, cas = ;
     scanf ("%d", &T);
     while (T--){
         cin >> s1 >> s2;
         int len1 = s1.size();
         int len2 = s2.size();
         Hash(s1, );
         Hash(s2, );
         string tmp1 = s1;
         string tmp2 = s2;
         reverse(tmp1.begin(), tmp1.end());
         reverse(tmp2.begin(), tmp2.end());
         exkmp(s2, len2, s1, len1, next, ex1);
         exkmp(tmp2, len2, tmp1, len1, next, ex2);
         int ans = -;
         for (int i = ; i < len1; i++){
             int t1 = ex1[i];
             int t2 = ex2[len1 - i - len2];
             int L1 = i+t1, R1 = i+len2-t2-;
             int L2 = t1, R2 = len2-t2-;
             bool ok1 = ((t1 == len2) || (R1 <= L1+));

             uLL p1 = (_hash[][R1-] - _hash[][L1] * bas[R1--L1]);
             uLL p2 = (_hash[][R2-] - _hash[][L2] * bas[R2--L2]);
             bool ok2 = (p1 == p2);
             if (ok1 || ok2){
                 ans = i;
                 break;
             }
         }
         printf("Case #%d: %d\n", cas++, ans);
     }
     return ;
 }