Strings in the Pocket 【manacher】

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=6012

题目描述

BaoBao has just found two strings $s=s_1s_2\dots s_n$ and $t=t_1t_2\dots t_n$ 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 $s_1s_2\dots s_{l-1}s_rs_{r-1}\dots s_{l+1}s_ls_{r+1}\dots s_{n-1}s_n$ .

In how many ways can BaoBao change to using the above operation exactly once? Let $(a,b)$ be an operation which reverses the substring $s_as_{a+1}\dots s_b$ , and $(c,d)$ be an operation which reverses the substring $s_cs_{c+1}\dots s_d$ . These two operations are considered different, if $a=\not c$ or $b=\not d$ .

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).

题目大意

输入两个长度相等字符串 $s,t$ ，将 $s$ 中的一个子串翻转，使得反转后的 $s$ 与 $t$ 相同，求有几种翻转方法

解题思路

若 $s与t完全相同$ ，那么可以翻转的方法数就等于将= $s$ 串中所有的回文串的数目
若 $s与t$ 不完全相同，那么从左往右找到第一个 $s[i]与t[i]$ 不同的位置 $l$ ，从右往左找到第一个 $s[i]与t[i]$ 不同的位置 $r$ ，如果 $s[l],s[l+1]...s[r]$ 翻转后与 $t[l],t[l+1]...t[r]$ 不同，那么不存在翻转方法，直接输出 $0$ ；如果 $s[l],s[l+1]...s[r]$ 翻转后与 $t[l],t[l+1]...t[r]$ 相同，那么以该子串往两边扩展，子串能扩展的最大长度就是可以翻转的方法数。

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <set>
using namespace std;
#define io ios::sync_with_stdio(0),cin.tie(0)
#define ms(arr) memset(arr,0,sizeof(arr))
#define mc(a,b) memcpy(a,b,sizeof(b))
#define inf 0x3f3f3f
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
typedef long long ll;
typedef unsigned long long ULL;
const int mod=1e9+7;
const int N=2e6+7;
char s[N],t[N],s_new[N<<1];
int p[N<<1];
int n;
void init()
{
    int len=strlen(s);
    s_new[0]='$';
    s_new[1]='#';
    for(int i=0; i<len; i++)
    {
        s_new[i*2+2]=s[i];
        s_new[i*2+3]='#';
    }
    s_new[2*len+2]='\0';
}
void manacher()
{
    int len=strlen(s_new);
    int id=-1;
    int mx=0;
    for(int i=1; i<len; i++)
    {
        if(i<mx) p[i]=min(p[id*2-i],mx-i);
        else p[i]=1;
        while(s_new[i+p[i]]==s_new[i-p[i]])
            p[i]++;
        if(mx<i+p[i])
        {
            mx=i+p[i];
            id=i;
        }
    }
}
int main()
{
//    fin;
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        s[0]='\0';
        t[0]='\0';
        scanf("%s",s);
        scanf("%s",t);
        int slen=strlen(s);
        int i,j;
        for(i=0;s[i]==t[i]&&i<slen;i++);
        if(i==slen)
        {
            init();
            manacher();
            ll ans=0;
            int len=strlen(s_new);
            for(int k=1;k<len;k++) ans+=(ll)(p[k]/2);
            printf("%lld\n",ans);
        }
        else{
            for(j=slen-1;s[j]==t[j]&&j>=0;j--);
            ll ans=1;
            for(int k=0;k<=j-i;k++)
            {
                if(s[i+k]!=t[j-k])
                {
                    ans=0;
                    break;
                }
            }
            if(ans==0) printf("0\n");
            else{
                i--;j++;
                while(i>=0&&j<slen&&s[i]==s[j])
                {
                    ans++;
                    i--;j++;
                }
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}