ZOJ 4110 Strings in the Pocket （马拉车+回文串）

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4110

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题目：

BaoBao has just found two strings  and  in his left pocket, where  indicates the -th character in string , and  indicates the -th character in string .

As BaoBao is bored, he decides to select a substring of  and reverse it. Formally speaking, he can select two integers  and  such that  and change the string to .

In how many ways can BaoBao change  to  using the above operation exactly once? Let  be an operation which reverses the substring , and  be an operation which reverses the substring . These two operations are considered different, if  or .

Input

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

The first line contains a string  (), while the second line contains another string  (). Both strings are composed of lower-cased English letters.

It's guaranteed that the sum of  of all test cases will not exceed .

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).

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题意：

存在S串和T串 要求对S串的一个子串做一次翻转操作可以得到T串的方案数

思路：

对子串分两种情况

第一种是S串和T串完全相同 可以的方案数就是S串中的所有的回文子串 因为S串长度为2e6 必须要用马拉车线性去处理出所有的回文子串

第二种是S串和T串有不同的部分 找出两个不同的字符最远的位置(l,r) 先判断S串的这个区间是否能通过翻转变成T串的区间 如果不可以直接输出0 如果可以 则向两侧同时延展寻找是否可以翻转

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代码：

#include <bits/stdc++.h>
 
using namespace std;
typedef long long ll;
const int maxn=2e6+;
int t,len[maxn*];
char S[maxn],T[maxn],s[maxn*];
 
int init(char *str){
    int n=strlen(str);
    for(int i=,j=;i<=*n;j++,i+=){
        s[i]='#';
        s[i+]=str[j];
    }
    s[]='$';
    s[*n+]='#';
    s[*n+]='@';
    s[*n+]='\n';
    return *n+;
}
 
void manacher(int n){
    int mx=,p=;
    for(int i=;i<=n;i++){
        if(mx>i) len[i]=min(mx-i,len[*p-i]);
        else len[i]=;
        while(s[i-len[i]]==s[i+len[i]]) len[i]++;
        if(len[i]+i>mx) mx=len[i]+i,p=i;
    }
}
 
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%s",S);
        scanf("%s",T);
        int Len=strlen(S),tot1=-,tot2=Len;
        for(int i=;i<Len;i++){
            if(S[i]!=T[i]){
                tot1=i;break;
            }
        }
        for(int i=Len-;i>=;i--){
            if(S[i]!=T[i]){
                tot2=i;break;
            }
        }
        if(tot1==-){
            int n=init(S);
            for(int i=;i<=n;i++) len[i]=;
            manacher(n);
            ll ans=;
            for(int i=;i<=n;i++) ans+=len[i]/;
            printf("%lld\n",ans);
            continue;
        }
        else{
            int tmp=;
            for(int i=tot1;i<=tot2;i++){
                if(S[i]!=T[tot2-(i-tot1)]){
                    tmp=;
                    break;
                }
            }
            if(tmp==){
                printf("0\n");
                continue;
            }
            else{
                ll ans=; tot1--; tot2++;
                while (tot1>= && tot2<Len && S[tot1]==T[tot2] && S[tot2]==T[tot1]){
                        tot1--; tot2++; ans++;
                }
                printf("%lld\n",ans);
            }
        }
    }
    return ;
}