hdu 2594 Simpsons’ Hidden Talents(扩展kmp)
Marge: Yeah, what is it?
Homer: Take me for example. I want to find out if I have a talent in politics, OK?
Marge: OK.
Homer: So I take some politician’s name, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.
The lengths of s1 and s2 will be at most 50000.
#include <cstdio>
#include <map>
#include <iostream>
#include<cstring>
#include<bits/stdc++.h>
#define ll long long int
#define M 6
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[]={,,,,,,,,,,,,};
int dir[][]={, ,, ,-, ,,-};
int dirs[][]={, ,, ,-, ,,-, -,- ,-, ,,- ,,};
const int inf=0x3f3f3f3f;
const ll mod=1e9+;
int nextt[];
int extend[];
void getnext(string s){
int po; int len=s.length();
nextt[]=len; //初始化next[0]
int pos=;
while(s[pos]==s[pos+]&&pos<len-) pos++;
nextt[]=pos; //计算next[1]
po=; //初始化po的位置
for(int i=;i<len;i++){
if(nextt[i-po]+i<po+nextt[po]) //第一种情况,可以直接得到next[i]的值
nextt[i]=nextt[i-po];
else{ //第二种情况,要继续匹配才能得到next[i]的值
int j=po+nextt[po]-i;
if(j<) j=;
while(i+j<len&&s[j]==s[i+j]) j++;
nextt[i]=j;
po=i;
}
//cout<<nextt[i]<<endl;
}
}
void exkmp(string a,string b){
int len1=a.length(); int len2=b.length();
getnext(a);
int po; int pos=;
while(a[pos]==b[pos]&&pos<len1&&pos<len2) pos++;
extend[]=pos;
po=;
for(int i=;i<len2;i++){
if(nextt[i-po]+i<extend[po]+po)
extend[i]=nextt[i-po];
else{
int j=po+extend[po]-i;
if(j<) j=;
while(i+j<len2&&a[j]==b[i+j]&&j<len1) j++;
extend[i]=j;
po=i;
}
}
}
int main(){
ios::sync_with_stdio(false);
string s1,s2;
while(cin>>s1>>s2){
exkmp(s1,s2);
int len1=s1.length(); int len2=s2.length();
int maxn=-inf;
int maxn_i;
for(int i=;i<len2;i++){
//cout<<extend[i]<<endl;
if(extend[i]+i==len2&&maxn<extend[i]){
maxn=extend[i];
maxn_i=i;
}
}
if(maxn==-inf) cout<<""<<endl;
else{ for(int i=maxn_i;i<maxn_i+maxn;i++)
cout<<s2[i];
cout<<" "<<maxn;
cout<<endl;
}
}
return ;
}
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