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+7;
const int MAXN = 1e5+10;
typedef unsigned long long uLL;
uLL _hash[2][MAXN];
string s1, s2;
void Hash(string s, int d){
    int len = s.size();
    memset(_hash[d], 0, sizeof (_hash[d]));
    for (int i = 0; i < len; i++){
        _hash[d][i] = (i ? _hash[d][i-1] : 0) * seed + s[i] - '0';
    }
}
void pre_kmp(string &s, int m, int next[]){
    next[0] = m;
    int j = 0;
    while (j + 1 < m && s[j] == s[j+1]){
        j++;
    }
    next[1] = j;
    int k = 1;
    for (int i = 2; i < m; i++){
        int p = next[k] + k - 1;
        int L = next[i-k];
        if (i + L < p + 1){
            next[i] = L;
        }else{
            j = max(0, p-i+1);
            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 = 0;
    while (j < len2 && j < len1 && str1[j] == str2[j]){
        j++;
    }
    extend[0] = j;
    int k = 0;
    for (int i = 1; i < len2; i++){
        int p = extend[k] + k - 1;
        int L = next[i-k];
        if (i + L < p + 1){
            extend[i] = L;
        }else{
            j = max(0, p-i+1);

            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[0] = 1;
    for (int i = 1; i < MAXN; i++){
        bas[i] = bas[i-1] * seed;
    }
}
int main(){

    //freopen("in.txt", "r", stdin);
    pre_solve();
    int T, cas = 1;
    scanf ("%d", &T);
    while (T--){
        cin >> s1 >> s2;
        int len1 = s1.size();
        int len2 = s2.size();
        Hash(s1, 0);
        Hash(s2, 1);
        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 = -1;
        for (int i = 0; i < len1; i++){
            int t1 = ex1[i];
            int t2 = ex2[len1 - i - len2];
            int L1 = i+t1, R1 = i+len2-t2-1;
            int L2 = t1, R2 = len2-t2-1;
            bool ok1 = ((t1 == len2) || (R1 <= L1+1));

            uLL p1 = (_hash[0][R1-1] - _hash[0][L1] * bas[R1-1-L1]);
            uLL p2 = (_hash[1][R2-1] - _hash[1][L2] * bas[R2-1-L2]);
            bool ok2 = (p1 == p2);
            if (ok1 || ok2){
                ans = i;
                break;
            }
        }
        printf("Case #%d: %d
", cas++, ans);
    }
    return 0;
}