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

Posted 2020-10-06 惜取少年时

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))  

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <vector>
  5 #include <set>
  6 #include <map>
  7 #include <string>
  8 #include <cstring>
  9 #include <stack>
 10 #include <queue>
 11 #include <cmath>
 12 #include <ctime>
 13 #include <bitset>
 14 #include <utility>
 15 #include <assert.h>
 16 using namespace std;
 17 #define rank rankk
 18 #define mp make_pair
 19 #define pb push_back
 20 #define xo(a,b) ((b)&1?(a):0)
 21 #define tm tmp
 22 //#define LL ll
 23 typedef unsigned long long ull;
 24 typedef pair<int,int> pii;
 25 typedef long long ll;
 26 typedef pair<ll,int> pli;
 27 typedef pair<ll,ll> pll;
 28 const int INF=0x3f3f3f3f;
 29 const ll INFF=0x3f3f3f3f3f3f3f3fll;
 30 const int MAX=1e5+5;
 31 //const ll MAXN=2e8;
 32 const int MAX_N=MAX;
 33 const ll MOD=998244353;
 34 //const long double pi=acos(-1.0);
 35 //const double eps=0.00000001;
 36 int gcd(int a,int b){return b?gcd(b,a%b):a;}
 37 template<typename T>inline T abs(T a) {return a>0?a:-a;}
 38 template<class T> inline
 39 void read(T& num) {
 40     bool start=false,neg=false;
 41     char c;
 42     num=0;
 43     while((c=getchar())!=EOF) {
 44         if(c==‘-‘) start=neg=true;
 45         else if(c>=‘0‘ && c<=‘9‘) {
 46             start=true;
 47             num=num*10+c-‘0‘;
 48         } else if(start) break;
 49     }
 50     if(neg) num=-num;
 51 }
 52 inline ll powMM(ll a,ll b,ll M){
 53     ll ret=1;
 54     a%=M;
 55 //    b%=M;
 56     while (b){
 57         if (b&1) ret=ret*a%M;
 58         b>>=1;
 59         a=a*a%M;
 60     }
 61     return ret;
 62 }
 63 void open()
 64 {
 65 //    freopen("1009.in","r",stdin);
 66     freopen("out.txt","w",stdout);
 67 }
 68 const int MAXN=100005;
 69 int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
 70 //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,
 71 //除s[n-1]外的所有s[i]都大于0，r[n-1]=0
 72 //函数结束以后结果放在sa数组中
 73 bool cmp(int *r,int a,int b,int l)
 74 {
 75     return r[a] == r[b] && r[a+l] == r[b+l];
 76 }
 77 void da(int str[],int sa[],int rank[],int height[],int n,int m)
 78 {
 79     str[n++]=0;
 80     int i, j, p, *x = t1, *y = t2;
 81     //第一轮基数排序，如果s的最大值很大，可改为快速排序
 82     for(i = 0;i < m;i++)c[i] = 0;
 83     for(i = 0;i < n;i++)c[x[i] = str[i]]++;
 84     for(i = 1;i < m;i++)c[i] += c[i-1];
 85     for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;
 86     for(j = 1;j <= n; j <<= 1)
 87     {
 88         p = 0;
 89         //直接利用sa数组排序第二关键字
 90         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
 91         for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
 92         //这样数组y保存的就是按照第二关键字排序的结果
 93         //基数排序第一关键字
 94         for(i = 0; i < m; i++)c[i] = 0;
 95         for(i = 0; i < n; i++)c[x[y[i]]]++;
 96         for(i = 1; i < m;i++)c[i] += c[i-1];
 97         for(i = n-1; i >= 0;i--)sa[--c[x[y[i]]]] = y[i];
 98         swap(x,y);
 99         //根据sa和x数组计算新的x数组 swap(x,y);
100         p = 1; x[sa[0]] = 0;
101         for(i = 1;i < n;i++)
102             x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
103         if(p >= n)break;
104         m = p;//下次基数排序的最大值
105     }
106     int k = 0; n--;
107     for(i = 0;i <= n;i++)rank[sa[i]] = i;
108     for(i = 0;i < n;i++)
109     {
110         if(k)k--;
111         j = sa[rank[i]-1]; while(str[i+k] == str[j+k])k++; height[rank[i]] = k;
112     }
113 }
114 int rank[MAXN],height[MAXN];
115 int RMQ[MAXN];
116 int mm[MAXN];
117 int best[20][MAXN];
118 void initRMQ(int n)
119 {
120     mm[0]=-1;
121     for(int i=1;i<=n;i++)
122     mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
123     for(int i=1;i<=n;i++)best[0][i]=i;
124     for(int i=1;i<=mm[n];i++)
125     for(int j=1;j+(1<<i)-1<=n;j++)
126     {
127         int a=best[i-1][j];
128         int b=best[i-1][j+(1<<(i-1))];
129         if(RMQ[a]<RMQ[b])best[i][j]=a; else best[i][j]=b;
130     }
131 }
132 int askRMQ(int a,int b)
133 {
134     int t; t=mm[b-a+1];
135     b-=(1<<t)-1;
136     a=best[t][a];b=best[t][b];
137     return RMQ[a]<RMQ[b]?a:b;
138 }
139 int lcp(int a,int b)
140 {
141     a=rank[a];b=rank[b];
142     if(a>b)swap(a,b);
143     return height[askRMQ(a+1,b)];
144 }
145 string sts[MAXN];
146 int st[MAXN],sa[MAXN];
147 
148 struct Trie {
149     bool isWord;
150     Trie* child[27];
151     Trie(bool isWord):isWord(isWord)
152     {
153         memset(child,0,sizeof(child));
154     }
155     void addWord(string &s)
156     {
157         Trie*cur =this;
158         for(char c: s)
159         {
160             Trie* next=cur->child[c-‘a‘+1];
161             if(next==nullptr)
162                 next=cur->child[c-‘a‘+1]=new Trie(false);
163             cur=next;
164         }
165         cur->isWord=true;
166     }
167     int checkstr(string s)
168     {
169         Trie*cur=this;int re=0;
170         for(char c:s)
171         {
172             Trie* next=cur->child[c-‘a‘+1];
173             if(next==nullptr)break;
174             else{cur=next;++re;}
175         }
176         return re;
177     }
178     ~Trie()
179     {
180         for(int i=0;i<27;++i)
181         {
182             if(child[i])
183                 delete child[i];
184         }
185     }
186 };
187 int t,n,m,x,y,an;
188 int main()
189 {
190     scanf("%d",&t);
191     while(t--)
192     {
193         scanf("%d",&n);Trie now(false);
194         for(int i=1;i<=n;i++){cin>>sts[i];now.addWord(sts[i]);}
195         scanf("%d",&m);
196         for(int i=1;i<=m;i++)
197         {
198             scanf("%d%d",&x,&y);int len1=sts[x].length(),len2=sts[y].length();
199             for(int i=0;i<len1;i++)st[i]=sts[x][i]-‘a‘+1;st[len1]=27;
200             for(int i=0;i<len2;i++)st[len1+1+i]=sts[y][i]-‘a‘+1;
201             da(st,sa,rank,height,len1+len2+1,28);an=0;
202             for(int i=1;i<=len1+len2+1;i++)
203             {
204                 int lo1=sa[i],lo2=sa[i-1];if(lo1>lo2)swap(lo1,lo2);
205                 if(height[i]>an&&lo1<len1&&lo2>len1)
206                 {
207                     int tem=now.checkstr(sts[x].substr(lo1,height[i]));
208                     if(tem>an)an=tem;
209                 }
210             }
211             printf("%d\n",an);
212         }
213     }
214 }