Strings in the Pocket

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

题意：反转某一个区间的子字符串，使得两个字符串相同

题解：

1、判断两个字符串是否完全相同；

2、完全相同则Manacher's Algorithm求回文字符串数量；

3、不完全相同则分别从头从尾至不相同得到区间[l，r],反转这个区间判断是否和另一个字符串相同；

4、3中不相同则ans==0；

5、3中相同则同时向两边扩张，条件是同时加1，并且相同

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=2000000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
char a[N],b[N];
int P[N<<1];
char str;
struct node{};
ll Manacher(string s)
{/*改造字符串*/string res="$#";for(int i=0;i<s.size();++i){res+=s[i];res+="#";}/*数组*/ll ans=0;int mi=0,right=0;   //mi为最大回文串对应的中心点，right为该回文串能达到的最右端的值for(int i=1;i<res.size();++i){P[i]=right>i ?min(P[2*mi-i],right-i):1;     //关键句，文中对这句以详细讲解while(res[i+P[i]]==res[i-P[i]])++P[i];if(right<i+P[i]){    //超过之前的最右端，则改变中心点和对应的最右端right=i+P[i];mi=i;}ans+=P[i]/2;}return ans;
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endifios::sync_with_stdio(false);cin.tie(0);//cout.tie(0);//scanf("%d",&t);cin>>t;while(t--){//scanf("%s%s",a,b);cin>>a>>b;int len=strlen(a);for(l=0;l<len&&a[l]==b[l];l++);for(r=len-1;r>=0&&a[r]==b[r];r--);if(l>r){cout<<Manacher(a)<<endl;}else{flag=1;for(int i=l;i<=r;i++){if(a[i]!=b[r-i+l]){flag=0;break;}}if(flag){ans=1;while(l-1>=0&&r+1<=len-1&&a[l-1]==a[r+1])ans++,l--,r++;cout<<ans<<endl;}else{cout<<0<<endl;}}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

C++版本二

原博客

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
using namespace std;
const int N=2e6+50;
char s1[N],s2[N],s[N*2];
int len[N*2];
int mi(int a,int b){if(a<b) return a;return b;
}
int main(){int t,i,n,k,p,ma;scanf("%d",&t);while(t--){scanf("%s%s",s1+1,s2+1);n=strlen(s1+1);k=1;for(i=1;i<=n;i++) if(s1[i]!=s2[i]) k=0;if(k){for(i=1;i<=n+1;i++){s[i*2]=s1[i];s[i*2-1]='*';}s[0]='#';p=ma=0;long long ans=0;for(i=1;i<=2*n;i++){if(ma>i) len[i]=mi(ma-i,len[2*p-i]);else len[i]=1;while(s[i+len[i]]==s[i-len[i]]) len[i]++;if(ma<len[i]+i){ma=len[i]+i;p=i;}ans+=len[i]/2;}printf("%lld\n",ans);}else{int l=1,r=n,ans=1;while(s1[l]==s2[l]) l++;while(s1[r]==s2[r]) r--;for(i=l;i<=r;i++) if(s1[i]!=s2[r-(i-l)]) ans=0;if(ans==0) printf("0\n");else{ma=1;while(l-ma>=1&&r+ma<=n&&s1[l-ma]==s1[r+ma]) ma++;printf("%d\n",ma);}}}return 0;
}

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