POJ 2217-Secretary（字尾陣列+高度陣列-最大公共子串長度）

kewlgrl發表於2017-04-22
陣列
Secretary
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 1382   Accepted: 568

Description

The basic condition of success of a political party, it is the good Election Programme. PSOS know about it, so they entrust the top secretary Juliet with this task. Because she wanted to make her work easier, she used her charm to talk round her friend Romeo to help her. Romeo is assistent of another political party and he was writing the programme some time ago. While writing the Programme for Juliet, he used some parts of his previous programme. When he gave the finished Programme to Juliet, they recognized that both programmes are too similar and that someone could notice it. They need to determine the longest part of text which is common to both programmes.

Input

At the first line there is a positive integer N stating the number of assignments to follow. Each assignment consists of exactly two lines of text, each of them contains at most 10000 characters. The end-of-line character is not considered to be a part of the text.

Output

Print a single line of text for each assignment. The line should contain the sentence "Nejdelsi spolecny retezec ma delku X." (The longest common part of text has X characters). Replace X with the length of the longest common substring of both texts.

Sample Input

2
Tady nejsou zadni mimozemstani.
Lide tady take nejsou.
Ja do lesa nepojedu.
V sobotu pojedeme na vylet.

Sample Output

Nejdelsi spolecny retezec ma delku 7.
Nejdelsi spolecny retezec ma delku 5.

Source

CTU FEE Local 1998


POJ 2774 買一送一~\(≧▽≦)/~啦啦啦


題目意思：

計算兩個字串的最大公共子串長度。


解題思路：

字尾陣列+高度陣列模板題。

在兩個串之間插入一個不會再串中出現的字元拼接成一個新串S'，計算高度陣列並求出其分屬於兩個不同串S和T時的最大值。



#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
#define MAXN 200100
int n,k;
string s,t;
int sa[MAXN],lcp[MAXN];
int rank[MAXN*2],tmp[MAXN*2];
void construct_lcp(string s,int sa[],int lcp[])
{
    int n=s.length();
    for(int i=0; i<=n; ++i)
        rank[sa[i]]=i;
    int h=0;
    lcp[0]=0;
    for(int i=0; i<n; ++i)
    {
        int j=sa[rank[i]-1];
        if(h>0) --h;
        for(; j+h<n&&i+h<n; ++h)
            if(s[j+h]!=s[i+h]) break;
        lcp[rank[i]-1]=h;
    }
}
bool compare_sa(int i,int j)//倍增法，比較rank
{
    if(rank[i]!=rank[j]) return rank[i]<rank[j];
    else
    {
        int ri=i+k<=n?rank[i+k] :-1;
        int rj=j+k<=n?rank[j+k] :-1;
        return ri<rj;
    }
}

void construct_sa(string s,int sa[])//計算s的字尾陣列
{
    for(int i=0; i<=n; ++i)//初始長度為1，rank為字元編碼
    {
        sa[i]=i;
        rank[i]=i<n?s[i] :-1;
    }
    for(k=1; k<=n; k*=2)//倍增法求字尾陣列
    {
        sort(sa,sa+n+1,compare_sa);
        tmp[sa[0]]=0;
        for(int i=1; i<=n; ++i)
            tmp[sa[i]]=tmp[sa[i-1]]+(compare_sa(sa[i-1],sa[i])?1:0);
        for(int i=0; i<=n; ++i)
            rank[i]=tmp[i];
    }
}
bool contain(string s,int sa[],string t)
{
    int a=0,b=s.length();
    while(b-a>1)
    {
        int c=(a+b)/2;
        if(s.compare(sa[c],t.length(),t)<0) a=c;
        else b=c;
    }
    return s.compare(sa[b],t.length(),t)==0;
}
void solve()
{
    int sl=s.length();
    s+='\0'+t;
    n=s.length();
    construct_sa(s,sa);
    construct_lcp(s,sa,lcp);
    int ans=0;
    for(int i=0; i<s.length(); ++i)
        if((sa[i]<sl)!=(sa[i+1]<sl))
            ans=max(ans,lcp[i]);
    cout<<"Nejdelsi spolecny retezec ma delku "<<ans<<"."<<endl;
}
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin>>T;
    string temp;
    getline(cin,temp);//吃掉T後面的回車
    while(T--)
    {
        getline(cin,s);//整行輸入
        getline(cin,t);
        solve();
    }
    return 0;
}


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