Strings in the Pocket

Time Limit: 1 Second Memory Limit: 65536 KB
BaoBao has just found two strings $s=s_1{s}_2\dots s_n$ and $t=t_1{t}_2\dots t_n$ in his left pocket, where $s_i$ indicates the $i$-th character in string , and $t_i$ indicates the $i$-th character in string $t$.

As BaoBao is bored, he decides to select a substring of and reverse it. Formally speaking, he can select two integers $l$ and $r$ such $1\le l\le r\le n$that and change the string to $s_1{s}_2\dots s_{l?1}{s}_r{s}_{r?1}\dots s_{l+1}{s}_l{s}_{r+1}\dots s_{n?1}{s}_n$.

In how many ways can BaoBao change to using the above operation exactly once? Let $(a,b)$ be an operation which reverses the substring $s_a{s}_{a+1}\dots s_b$, and $(c,d)$ be an operation which reverses the substring $s_c{s}_{c+1}\dots s_d$. These two operations are considered different, if $a=?c$ or $b=?d$.

Input

There are multiple test cases. The first line of the input contains an integer $T$, indicating the number of test cases. For each test case:

The first line contains a string $s(1\le |s|\le 2\times 10^6)$, while the second line contains another string $t(|t|=|s|)$. Both strings are composed of lower-cased English letters.

It’s guaranteed that the sum of $|s|$ of all test cases will not exceed $2\times 10^7$.

Output

For each test case output one line containing one integer, indicating the answer.

Sample Input

2
abcbcdcbd
abcdcbcbd
abc
abc

Sample Output

3
3
Hint
For the first sample test case, BaoBao can do one of the following three operations: (2, 8), (3, 7) or (4, 6).

For the second sample test case, BaoBao can do one of the following three operations: (1, 1), (2, 2) or (3, 3).

題目大意：

先輸入一個整數$t$，代表共有$t$組測試樣例，對于每一組測試樣例，輸入兩個字符串$s_1,s_2$，現在僅能將字符串$s_1$中的某個字串翻轉$s_1$與$s_2$相等，問共有幾中反轉方法。

解題思路：

此題可以將題目分為兩種情況，一種是字符串$s_1$與字符串$s_2$完全相同，對于這種情況，可以通過manacher來計算字符串$s_i$中每個回文串的長度，將長度求和即為總的方案數，對于$s_1$與$s_2$不等的情況，可以先從左往右找到$s_1$與$s_2$第一個不同的位置$l$，同理，從右往左找到兩字符串第一次不同的位置$r$，先判斷$s_1$的字串$s_l{s}_{l+1}\dots s_{r?1}{s}_r$反轉后是否會使$s_1$與$s_2$完全相同，不會的話證明無法在反轉一次的情況下使兩字符串相同，輸出$0$，否則則以此字串為邊界同時往兩邊擴展，判斷字串長度是否能延伸，記錄輸出即可。

代碼

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
string s1,s2,s1_new;
int p[6004000];
int get_newstring() {s1_new.clear();int len=s1.size();s1_new+='$';for(int i=0;i<len;i++) {s1_new+='#';s1_new+=s1[i];}s1_new+='#';return s1_new.size();
}
ll manacher() {int len=get_newstring();fill(p,p+len,0);int id=0,mx=0;ll ans=0;for(int i=0;i<len;i++) {if(i<mx) p[i]=min(p[2*id-i],mx-i);elsep[i]=1;while(s1_new[i-p[i]]==s1_new[i+p[i]]) {p[i]++;}if(p[i]+i>mx) {mx=i+p[i];id=i;}ans+=(ll)(p[i]/2);}return ans;
}
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int t;cin>>t;while(t--) {cin>>s1>>s2;ll ans=0;if(s1==s2) {ans=manacher();}else {int l=0,r=s1.size()-1;while(s1[l]==s2[l]) l++;while(s1[r]==s2[r]) r--;bool ju=false;int t1=l,t2=r;while(s1[t1]==s2[t2]) {if(t1==r&&t2==l) break;t1++;t2--;}if(t2==l&&t1==r) {ju=true;}else ju=false;if(l==r) ju=false;if(ju==true) {ans=1;l--,r++;while(l>=0&&r<s1.size()&&s1[l]==s2[r]&&s1[r]==s2[l]) {ans++;l--;r++;}}else ans=0;}cout<<ans<<endl;}return 0;
}