hdu 2328 Corporate Identity(kmp)

Problem Description

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a help with their new identity. IBM do not want to change their existing logos and trademarks completely, because their customers are used to the old ones. Therefore, ACM will only change existing trademarks instead of creating new ones.

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.

Input

The input contains several tasks. Each task begins with a line containing a positive integer N, the number of trademarks (2 ≤ N ≤ 4000). The number is followed by N lines, each containing one trademark. Trademarks will be composed only from lowercase letters, the length of each trademark will be at least 1 and at most 200 characters.

After the last trademark, the next task begins. The last task is followed by a line containing zero.

Output

For each task, output a single line containing the longest string contained as a substring in all trademarks. If there are several strings of the same length, print the one that is lexicographically smallest. If there is no such non-empty string, output the words “IDENTITY LOST” instead.

Sample Input

3

aabbaabb

abbababb

bbbbbabb

2

xyz

abc

0

Sample Output

abb

IDENTITY LOST

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[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
int n;
string s[4007];
int nextt[500];
void getnext(string s){
    nextt[1]=0;
    int len=s.length();
    for(int i=2,j=0;i<=len;i++){
        while(j>0&&s[i-1]!=s[j]) j=nextt[j];
        if(s[i-1]==s[j]) j++;
        nextt[i]=j;
    }
}
bool kmp(string t,string p){
    int len1=p.length();
    int len2=t.length();
    for(int i=1,j=0;i<=len1;i++){
        while(j>0&&p[i-1]!=t[j]) j=nextt[j];
        if(p[i-1]==t[j]) j++;
        if(j==len2) return true;
    }
    return false;
}
int main(){
    ios::sync_with_stdio(false);
    while(cin>>n&&n){
        for(int i=0;i<n;i++)
            cin>>s[i];
        int len=s[0].length();
        string ans="";
        for(int i=len;i>=1;i--){
            for(int j=0;j<=len-i;j++){
                string temp=s[0].substr(j,i);
                getnext(temp);
                bool f=1;
                for(int k=0;k<n;k++)
                    if(!kmp(temp,s[k])){
                        f=0;
                        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()<1) cout<<"IDENTITY LOST"<<endl;
        else cout<<ans<<endl;

    }
    return 0;
}    

原文地址：https://www.cnblogs.com/wmj6/p/10504938.html