POJ 2217 Secretary

Secretary

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2217
64-bit integer IO format: %lld      Java class name: Main

 

 

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

 

解题：后缀数组。将两个串S，T链接起来。然后求lcp.sa[i]和sa[i+1]不同时在S或者不同时在T中，求满足这样条件的最大lcp。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 300000;
int n,k,_rank[maxn],sa[maxn],lcp[maxn],tmp[maxn];
char ss[100010];
bool cmp_sa(int i,int j){
    if(_rank[i] != _rank[j]) return _rank[i] < _rank[j];
    int ri = i+k <= n ? _rank[i+k]:-1;
    int rj = j+k <= n ? _rank[j+k]:-1;
    return ri < rj;
}
void construct_sa(char *S,int *sa){
    for(int i = 0; i <= n; i++){
        sa[i] = i;
        _rank[i] = i < n ? S[i]:-1;
    }
    for(k = 1; k <= n; k <<= 1){
        sort(sa,sa+n+1,cmp_sa);
        tmp[sa[0]] = 0;
        for(int i = 1; i <= n; i++)
            tmp[sa[i]] = tmp[sa[i-1]] + cmp_sa(sa[i-1],sa[i]);
        for(int i = 0; i <= n; i++)
            _rank[i] = tmp[i];
    }
}
void construct_lcp(char *S,int *lcp){
    for(int i = 0; i <= n; i++) _rank[sa[i]] = i;
    int h = lcp[0] = 0;
    for(int i = 0; i < n; i++){
        if(h) h--;
        for(int j = sa[_rank[i]-1]; i+h < n && j+h < n && S[i+h] == S[j+h]; ++h);
        lcp[_rank[i]-1] = h;
    }
}
int main() {
    int t,slen;
    scanf("%d",&t);
    getchar();
    while(t--){
        gets(ss);
        slen = strlen(ss);
        ss[slen] = '#';
        gets(ss+slen+1);
        n = strlen(ss);
        memset(sa,0,sizeof(sa));
        memset(lcp,0,sizeof(lcp));
        construct_sa(ss,sa);
        construct_lcp(ss,lcp);
        int ans = 0;
        for(int i = 0; i < n; i++)
            if((sa[i] < slen) != (sa[i+1] < slen)) ans = max(ans,lcp[i]);
        printf("Nejdelsi spolecny retezec ma delku %d.
",ans);
    }
    return 0;
}