（后缀数组/Trie）HDU 6138-Fleet of the Eternal Throne

Fleet of the Eternal Throne

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 838    Accepted Submission(s): 393

Problem Description

> The Eternal Fleet was built many centuries ago before the time of Valkorion by an unknown race on the planet of Iokath. The fate of the Fleet's builders is unknown but their legacy would live on. Its first known action was in the annihilation of all life in Wild Space. It spread across Wild Space and conquered almost every inhabited world within the region, including Zakuul. They were finally defeated by a mysterious vessel known as the Gravestone, a massive alien warship that countered the Eternal Fleet's might. Outfitted with specialized weapons designed to take out multiple targets at once, the Gravestone destroyed whole sections of the fleet with a single shot. The Eternal Fleet was finally defeated over Zakuul, where it was deactivated and hidden away. The Gravestone landed in the swamps of Zakuul, where the crew scuttled it and hid it away.
>
> — Wookieepedia

The major defeat of the Eternal Fleet is the connected defensive network. Though being effective in defensing a large fleet, it finally led to a chain-reaction and was destroyed by the Gravestone. Therefore, when the next generation of Eternal Fleet is built, you are asked to check the risk of the chain reaction.

The battleships of the Eternal Fleet are placed on a 2D plane of n rows. Each row is an array of battleships. The type of a battleship is denoted by an English lowercase alphabet. In other words, each row can be treated as a string. Below lists a possible configuration of the Eternal Fleet.


aa
bbbaaa
abbaababa
abba


If in the x-th row and the y-th row, there exists a consecutive segment of battleships that looks identical in both rows (i.e., a common substring of the x-th row and y-th row), at the same time the substring is a prefix of any other row (can be the x-th or the y-th row), the Eternal Fleet will have a risk of causing chain reaction.

Given a query (x, y), you should find the longest substring that have a risk of causing chain reaction.

 

Input

The first line of the input contains an integer T, denoting the number of test cases. 

For each test cases, the first line contains integer n (n≤105).

There are n lines following, each has a string consisting of lower case letters denoting the battleships in the row. The total length of the strings will not exceed 105.

And an integer m (1≤m≤100) is following, representing the number of queries. 

For each of the following m lines, there are two integers x,y, denoting the query.

 

Output

You should output the answers for the queries, one integer per line.

 

Sample Input

1 3 aaa baaa caaa 2 2 3 1 2

 

Sample Output

3 3

O（总长）对所有串建立Trie树，m次询问，每次将两个串拼接起来，中间用非小写字母的符号隔开，求出后缀数组中的height，扫一遍height数组，遇到值比当前ans大并且对应sa[i],sa[i-1]分别在len1两侧（即两个后缀分别起始于第一个串、第二个串）时在Trie树中跑一下，更新ans。总时间复杂度 (总长）+m*len*log(len)+m*len =O（m*len*log(len))  

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <bitset>
#include <utility>
#include <assert.h>
using namespace std;
#define rank rankk
#define mp make_pair
#define pb push_back
#define xo(a,b) ((b)&1?(a):0)
#define tm tmp
//#define LL ll
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef long long ll;
typedef pair<ll,int> pli;
typedef pair<ll,ll> pll;
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3fll;
const int MAX=1e5+5;
//const ll MAXN=2e8;
const int MAX_N=MAX;
const ll MOD=998244353;
//const long double pi=acos(-1.0);
//const double eps=0.00000001;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
template<typename T>inline T abs(T a) {return a>0?a:-a;}
template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
inline ll powMM(ll a,ll b,ll M){
    ll ret=1;
    a%=M;
//    b%=M;
    while (b){
        if (b&1) ret=ret*a%M;
        b>>=1;
        a=a*a%M;
    }
    return ret;
}
void open()
{
//    freopen("1009.in","r",stdin);
    freopen("out.txt","w",stdout);
}
const int MAXN=100005;
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
bool cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int str[],int sa[],int rank[],int height[],int n,int m)
{
    str[n++]=0;
    int i, j, p, *x = t1, *y = t2;
    //第一轮基数排序，如果s的最大值很大，可改为快速排序
    for(i = 0;i < m;i++)c[i] = 0;
    for(i = 0;i < n;i++)c[x[i] = str[i]]++;
    for(i = 1;i < m;i++)c[i] += c[i-1];
    for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;
    for(j = 1;j <= n; j <<= 1)
    {
        p = 0;
        //直接利用sa数组排序第二关键字
        for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
        for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
        //这样数组y保存的就是按照第二关键字排序的结果
        //基数排序第一关键字
        for(i = 0; i < m; i++)c[i] = 0;
        for(i = 0; i < n; i++)c[x[y[i]]]++;
        for(i = 1; i < m;i++)c[i] += c[i-1];
        for(i = n-1; i >= 0;i--)sa[--c[x[y[i]]]] = y[i];
        swap(x,y);
        //根据sa和x数组计算新的x数组 swap(x,y);
        p = 1; x[sa[0]] = 0;
        for(i = 1;i < n;i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
        if(p >= n)break;
        m = p;//下次基数排序的最大值
    }
    int k = 0; n--;
    for(i = 0;i <= n;i++)rank[sa[i]] = i;
    for(i = 0;i < n;i++)
    {
        if(k)k--;
        j = sa[rank[i]-1]; while(str[i+k] == str[j+k])k++; height[rank[i]] = k;
    }
}
int rank[MAXN],height[MAXN];
int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];
void initRMQ(int n)
{
    mm[0]=-1;
    for(int i=1;i<=n;i++)
    mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    for(int i=1;i<=n;i++)best[0][i]=i;
    for(int i=1;i<=mm[n];i++)
    for(int j=1;j+(1<<i)-1<=n;j++)
    {
        int a=best[i-1][j];
        int b=best[i-1][j+(1<<(i-1))];
        if(RMQ[a]<RMQ[b])best[i][j]=a; else best[i][j]=b;
    }
}
int askRMQ(int a,int b)
{
    int t; t=mm[b-a+1];
    b-=(1<<t)-1;
    a=best[t][a];b=best[t][b];
    return RMQ[a]<RMQ[b]?a:b;
}
int lcp(int a,int b)
{
    a=rank[a];b=rank[b];
    if(a>b)swap(a,b);
    return height[askRMQ(a+1,b)];
}
string sts[MAXN];
int st[MAXN],sa[MAXN];

struct Trie {
    bool isWord;
    Trie* child[27];
    Trie(bool isWord):isWord(isWord)
    {
        memset(child,0,sizeof(child));
    }
    void addWord(string &s)
    {
        Trie*cur =this;
        for(char c: s)
        {
            Trie* next=cur->child[c-'a'+1];
            if(next==nullptr)
                next=cur->child[c-'a'+1]=new Trie(false);
            cur=next;
        }
        cur->isWord=true;
    }
    int checkstr(string s)
    {
        Trie*cur=this;int re=0;
        for(char c:s)
        {
            Trie* next=cur->child[c-'a'+1];
            if(next==nullptr)break;
            else{cur=next;++re;}
        }
        return re;
    }
    ~Trie()
    {
        for(int i=0;i<27;++i)
        {
            if(child[i])
                delete child[i];
        }
    }
};
int t,n,m,x,y,an;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);Trie now(false);
        for(int i=1;i<=n;i++){cin>>sts[i];now.addWord(sts[i]);}
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);int len1=sts[x].length(),len2=sts[y].length();
            for(int i=0;i<len1;i++)st[i]=sts[x][i]-'a'+1;st[len1]=27;
            for(int i=0;i<len2;i++)st[len1+1+i]=sts[y][i]-'a'+1;
            da(st,sa,rank,height,len1+len2+1,28);an=0;
            for(int i=1;i<=len1+len2+1;i++)
            {
                int lo1=sa[i],lo2=sa[i-1];if(lo1>lo2)swap(lo1,lo2);
                if(height[i]>an&&lo1<len1&&lo2>len1)
                {
                    int tem=now.checkstr(sts[x].substr(lo1,height[i]));
                    if(tem>an)an=tem;
                }
            }
            printf("%d
",an);
        }
    }
}