题意:
给字符串S和T,要求使S变为T
操作:对S的区间(l,r)反转,方法数
两个字符串相等的情况,MANACHER求回文子串的个数
不相等的话,找到s和t不相同的位置l,r,然后判断反转后能不能相等,不相等就不能翻转,相等的话,就往两边拓展答案。
#include
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
struct MANACHER{
#define STRSIZE 2010000
char str[STRSIZE*2+5];
int radius[STRSIZE*2+5];
int str_len;
MANACHER(){}
MANACHER(char *tar){solver(tar);}
int solver(char *tar); //执行MANACHER算法
int getMaxPalSize(); //最长回文串长度
long long getPalNum(); //不同回文串个数
};
int MANACHER::solver(char *tar){
//(1)construct new string
int tar_len=strlen(tar);
str_len=tar_len*2+2;
str[0]='~';
str[1]='!';
for (int i=0;i<tar_len;i++){
str[i*2+2]=tar[i];
str[i*2+3]='!';
}
str[str_len]=0;
//(2)calculate radius
int max_idx=-1;
int center=-1;
for (int i=0;i<str_len;i++){
radius[i]=1;
if (max_idx>i)
radius[i]=min(max_idx-i+1,radius[center*2-i]);
while(str[i+radius[i]]==str[i-radius[i]]) radius[i]++;
//update max_idx
if (max_idx<i+radius[i]-1){
max_idx=i+radius[i]-1;
center=i;
}
}
return 0;
}
int MANACHER::getMaxPalSize(){
int ans=0;
for (int i=1;i<str_len;i++)
ans=max(ans,radius[i]-1);
return ans;
}
long long MANACHER::getPalNum(){
long long ans=0;
for (int i=1;i<str_len;i++){
ans+=radius[i]/2;
}
return ans;
}
MANACHER x;
const int maxn = 2e6+100;
int T;
char str1[maxn],str2[maxn];
int main(int argc, char const *argv[])
{
// ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
scanf("%d",&T);
while(T--)
{
// cin >> str1 >> str2;
scanf("%s%s",str1,str2);
if(strcmp(str1,str2) == 0)
{
x.solver(str1);
printf("%lld\n",x.getPalNum());
}
else
{
ll ans = 0;
int le = strlen(str1);
int l = 0;
int r = 0;
rep(i,0,le)
{
if(str1[i] != str2[i])
{
l = i;
break;
}
}
per(i,0,le)
{
if(str1[i] != str2[i])
{
r = i;
break;
}
}
int j = r;
int f = 0;
rep(i,l,r+1)
{
if(str1[i] != str2[j])
{
f = 1;
break;
}
j--;
}
if(f)
{
printf("0\n");
}
else
{
ans = 1;
while(l-1 >= 0 && r+1 < le &&str1[l-1] == str2[r+1])
{
ans++;
l--;
r++;
}
printf("%lld\n",ans);
}
}
}
return 0;
}