ZOJ-4110 Strings in the Pocket 2019浙江省省赛

题目链接

Strings in the Pocket

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Time Limit: 1 Second      Memory Limit: 65536 KB

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BaoBao has just found two strings s=s1s2…sn and t=t1t2…tn in his left pocket, where si indicates the i-th character in string s, and ti indicates the i-th character in string t.

As BaoBao is bored, he decides to select a substring of s and reverse it. Formally speaking, he can select two integers l and r such that 1≤l≤r≤n and change the string to s1s2…sl−1srsr−1…sl+1slsr+1…sn−1sn.

In how many ways can BaoBao change s to t using the above operation exactly once? Let (a,b) be an operation which reverses the substring sasa+1…sb, and (c,d) be an operation which reverses the substring scsc+1…sd. 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≤|s|≤2×106), 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 2×107.

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).

 

题意：给你两字符串a，b问你把a的一段区间反转能得到b的次数。

当两个字符串相等时，直接用manacher求个数。

如果两个字符串不相等时，找到不相等的前后两个位置，然后往两边更新就好了

代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=2e6+5;
char a[maxn],b[maxn];
int n,Len[maxn],Lentwo[maxn],T;
ll manacher()//manacher算法 
{
	int l=0,r=0,x;
	ll ans=0;
	for(int i=1; i<=n; i++)//以奇数个字符为中心找 
	{
		if(i>r)x=1;
		else x=min(Len[l+r-i],r-i);
		while(i-x>=1&&i+x<=n&&a[i-x]==a[i+x])x++;
		Len[i]=x;
		ans+=x;
		if(i+x-1>r)
		{
			r=i+x-1;
			l=i-x+1;
		}
	}
	l=r=0;
	for(int i=1; i<=n; i++)//以偶数个字符为中心找 
	{
		if(i>r)x=0;
		else x=min(Lentwo[l+r-i+1],r-i+1);
		while(i-x-1>=1&&i+x<=n&&a[i-x-1]==a[i+x])x++;
		Lentwo[i]=x;
		ans+=x;
		if(i+x>=r)
		{
			l=i-x;
			r=i+x-1;
		}
	}
	return ans;
}
int main()
{
	cin>>T;
	while(T--)
	{
		scanf("%s",a+1);
		scanf("%s",b+1);
		n=strlen(a+1);
		int x=1,y=n;
		for(x; x<=n; x++)
			if(a[x]!=b[x])
				break;
		for(y; y>=1; y--)
			if(a[y]!=b[y])
				break;
		if(x==y)
		{
			printf("0\n");
		}
		else if(x<=n)
		{
			int ans=1;
			for(int i=x; i<=y; i++)
			{
				if(a[i]!=b[x+y-i])
				{
					ans=0;
					break;
				}
			}
			if(ans)
			{
				x--;
				y++;
				while(x>=1&&y<=n&&a[x]==b[y]&&a[y]==b[x])
				{
					ans++;
					x--;
					y++;
				}
			}
			printf("%d\n",ans);
		}
		else
		{
			printf("%lld\n",manacher());
		}
	}
}