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\n", cas++, ans);
}
return 0;
}