hdu 2328 Corporate Identity(kmp)
After several other proposals, it was decided to take all existing trademarks and find the longest common sequence of letters that is contained in all of them. This sequence will be graphically emphasized to form a new logo. Then, the old trademarks may still be used while showing the new identity.
Your task is to find such a sequence.
After the last trademark, the next task begins. The last task is followed by a line containing zero.
题意&思路:hdu 1238 数据的加强版 需要注意优化点已在代码处标记
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
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 n;
string s[];
int nextt[];
void getnext(string s){
nextt[]=;
int len=s.length();
for(int i=,j=;i<=len;i++){
while(j>&&s[i-]!=s[j]) j=nextt[j];
if(s[i-]==s[j]) j++;
nextt[i]=j;
}
}
bool kmp(string t,string p){
int len1=p.length();
int len2=t.length();
for(int i=,j=;i<=len1;i++){
while(j>&&p[i-]!=t[j]) j=nextt[j];
if(p[i-]==t[j]) j++;
if(j==len2) return true;
}
return false;
}
int main(){
ios::sync_with_stdio(false);
while(cin>>n&&n){
for(int i=;i<n;i++)
cin>>s[i];
int len=s[].length();
string ans="";
for(int i=len;i>=;i--){
for(int j=;j<=len-i;j++){
string temp=s[].substr(j,i);
getnext(temp);
bool f=;
for(int k=;k<n;k++)
if(!kmp(temp,s[k])){
f=;
break; //碰到假就弹出
}
if(f){
if(ans.size()<temp.size()) ans=temp;
else if(ans.size()==temp.size()) ans=min(ans,temp);
}
}
}
if(ans.size()<) cout<<"IDENTITY LOST"<<endl;
else cout<<ans<<endl; }
return ;
}
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