題目大意：給定N個串，求每個串至少出現兩次的最長子串。

解題思路：每個字串至少出現兩次且不可重疊的最長子串：二分列舉長度後在同一分組中對每一個字串保留一個最小的位置和一個最大的位置，最後檢視是否每個串在同一組中都有至少兩個字尾，並且字尾的座標差大於列舉的長度。

POJ Problem Set (classical)

220. Relevant Phrases of Annihilation

Problem code: PHRASES

You are the King of Byteland. Your agents have just intercepted a batch of encrypted enemy messages concerning the date of the planned attack on your island. You immedietaly send for the Bytelandian Cryptographer, but he is currently busy eating popcorn and claims that he may only decrypt the most important part of the text (since the rest would be a waste of his time). You decide to select the fragment of the text which the enemy has strongly emphasised, evidently regarding it as the most important. So, you are looking for a fragment of text which appears in all the messages disjointly at least twice. Since you are not overfond of the cryptographer, try to make this fragment as long as possible.

Input

The first line of input contains a single positive integer t<=10, the number of test cases. t test cases follow.Each test case begins with integer n (n<=10), the number of messages. The next n lines contain the messages, consisting only of between 2 and 10000 characters 'a'-'z', possibly with some additional trailing white space which should be ignored.

Output

For each test case output the length of longest string which appears disjointly at least twice in all of the messages.

Example

Input:
1
4
abbabba
dabddkababa
bacaba
baba

Output:
2

(in the example above, the longest substring which fulfills the requirements is 'ba')

---

Added by:	Adrian Kosowski
Date:	2004-10-11
Time limit:	9s
Source limit:	50000B
Memory limit:	1536MB
Cluster:	Cube (Intel Pentium G860 3GHz)
Languages:	All except: NODEJS PERL 6 SCM chicken VB.net
Resource:	DASM Programming League 2004 (problemset 1)

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;



inline int read()
{
    char ch;
    bool flag = false;
    int a = 0;
    while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
    if(ch != '-')
    {
        a *= 10;
        a += ch - '0';
    }
    else
    {
        flag = true;
    }
    while(((ch = getchar()) >= '0') && (ch <= '9'))
    {
        a *= 10;
        a += ch - '0';
    }
    if(flag)
    {
        a = -a;
    }
    return a;
}
void write(int a)
{
    if(a < 0)
    {
        putchar('-');
        a = -a;
    }
    if(a >= 10)
    {
        write(a / 10);
    }
    putchar(a % 10 + '0');
}

const int maxn = 200010;



int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];

int cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}


void da(int *r, int *sa, int n, int m)
{
    int i, j, p, *x = wa, *y = wb;
    for(i = 0; i < m; i++) ws1[i] = 0;
    for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
    for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
    for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;
    for(j = 1, p = 1; p < n; j <<= 1, m = p)
    {
        for(p = 0, i = n-j; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++)
            if(sa[i] >= j) y[p++] = sa[i]-j;
        for(i = 0; i < n; i++) wv[i] = x[y[i]];
        for(i = 0; i < m; i++) ws1[i] = 0;
        for(i = 0; i < n; i++) ws1[wv[i]]++;
        for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
        for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
        for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
    }
    return ;
}

int rank[maxn], height[maxn];

void calheight(int *r, int *sa, int n)
{
    int i, j, k = 0;
    for(i = 1; i <= n; i++) rank[sa[i]] = i;
    for(i = 0; i < n; height[rank[i++]] = k)
        for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
    return;
}

char str1[maxn], str2[maxn];
int seq[maxn];

int hash[maxn];

char str[110][1010];

int vis[110];

int Find(int x)
{
    int s = hash[x];
    for(int i = 0; i < s; i++)
    {
        int len = strlen(str[i]);
        x -= len;
    }
    x -= s;
    return x;
}

int pos[20][2];

bool judge(int mid, int n, int m)
{
    for(int i = 2; i <= n; i++)
    {
        memset(vis, 0, sizeof(vis));
        vis[hash[sa[i-1]]]++;
        pos[hash[sa[i-1]]][0] = pos[hash[sa[i-1]]][1] = sa[i-1];
        while(mid <= height[i])
        {
            int xp = hash[sa[i]];
            if(!vis[xp])
            {
                vis[xp]++;
                pos[xp][1] = pos[xp][0] = sa[i];
                i++;
                continue;
            }
            vis[xp]++;
            pos[xp][0] = max(pos[xp][0], sa[i]);
            pos[xp][1] = min(pos[xp][1], sa[i]);
            i++;
        }
        int k;
        for(k = 0; k < m; k++)
        {
            if(vis[k] < 2 || pos[k][0]-pos[k][1] < mid)
                break;
        }
        if(k == m) return true;
    }

    return false;
}

void Del(int n, int len, int m)
{
    int l = 1;
    int r = len/2;
    int xans = 0;
    while(l <= r)
    {
        int mid = (l+r)>>1;
        if(judge(mid, n, m))
        {
            l = mid+1;
            xans = mid;
        }
        else r = mid-1;
    }
    cout<<xans<<endl;
}


int main()
{
    int T;
    cin >>T;
    int n;
    while(T--)
    {
        scanf("%d", &n);
        memset(hash, -1, sizeof(hash));
        int ans = 0;
        int Min = maxn;
        for(int i = 0; i < n; i++)
        {
            scanf("%s",str[i]);
            int len = strlen(str[i]);
            Min = min(Min, len);
            for(int j = 0; j < len; j++)
            {
                seq[ans] = str[i][j];
                hash[ans++] = i;
            }
            seq[ans++] = 200+i;
        }
        seq[ans] = 0;
        da(seq, sa, ans+1, 310);
        calheight(seq, sa, ans);
        Del(ans, Min, n);
    }
    return 0;
}