（後綴數組/Trie）HDU 6138-Fleet of the Eternal Throne
阿新 • • 發佈：2017-09-12
cto ould itl nal pan iostream row onf map
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> — 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
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
Given a query (x, y), you should find the longest substring that have a risk of causing chain reaction.
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.
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
>
> — 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
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
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
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（總長）對所有串建立Trie樹，m次詢問，每次將兩個串拼接起來，中間用非小寫字母的符號隔開，求出後綴數組中的height，掃一遍height數組，遇到值比當前ans大並且對應sa[i],sa[i-1]分別在len1兩側（即兩個後綴分別起始於第一個串、第二個串）時在Trie樹中跑一下，更新ans。總時間復雜度 (總長）+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數組需要的中間變量，不需要賦值 70 //待排序的字符串放在s數組中，從s[0]到s[n-1],長度為n,且最大值小於m, 71 //除s[n-1]外的所有s[i]都大於0，r[n-1]=0 72 //函數結束以後結果放在sa數組中 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 //第一輪基數排序，如果s的最大值很大，可改為快速排序 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數組排序第二關鍵字 90 for(i = n-j; i < n; i++)y[p++] = i;//後面的j個數第二關鍵字為空的最小 91 for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j; 92 //這樣數組y保存的就是按照第二關鍵字排序的結果 93 //基數排序第一關鍵字 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 //根據sa和x數組計算新的x數組 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;//下次基數排序的最大值 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 }
（後綴數組/Trie）HDU 6138-Fleet of the Eternal Throne