HDU 5343 MZL's Circle Zhou

MZL's Circle Zhou

Time Limit: 1000ms

Memory Limit: 131072KB

This problem will be judged on HDU. Original ID: 5343
64-bit integer IO format: %I64d Java class name: Main

MZL's Circle Zhou is good at solving some counting problems. One day, he comes up with a counting problem:
You are given two strings a,b which consist of only lowercase English letters. You can subtract a substring x (maybe empty) from string a and a substring y (also maybe empty) from string b, and then connect them as x+y with x at the front and y at the back. In this way, a series of new strings can be obtained.
The question is how many different new strings can be obtained in this way.
Two strings are different, if and only if they have different lengths or there exists an integer i such that the two strings have different characters at position i.

Input
The first line of the input is a single integer T (T≤5), indicating the number of testcases.
For each test case, there are two lines, the first line is string a, and the second line is string b. 1<=|a|,|b|<=90000.

Output

For each test case, output one line, a single integer indicating the answer.

Sample Input

2

acbcc

cccabc

bbbabbababbababbaaaabbbbabbaaaabaabbabbabbbaaabaab

abbaabbabbaaaabbbaababbabbabababaaaaabbaabbaabbaab

Sample Output

135

557539

Source

2015 Multi-University Training Contest 5

解题：后缀自动机

A、B两串分别建两个SAM处理，先将B串翻转，利用SAM(B)求以字符c结尾（开头）的子串个数。

对于SAM(A)中节点x，若x不能接受字符c，则将B串中以c开头的子串“接”在其后，构成Asub+Bsub的形式。

 #include <bits/stdc++.h>

 using namespace std;

 typedef unsigned long long ULL;

 const int maxn = ;

 struct node{

     int son[],f,len,alpha;

     void init(){

         memset(son,-,sizeof son);

         f = -;

         len = ;

     }

 };

 struct SAM{

     node e[maxn];

     int tot,last;

     int newnode(int len = ,int alpha = ){

         e[tot].init();

         e[tot].len = len;

         e[tot].alpha = alpha;

         return tot++;

     }

     void init(){

         tot = last = ;

         newnode();

     }

     void add(int c){

         int p = last,np = newnode(e[p].len + ,c);

         while(p != - && e[p].son[c] == -){

             e[p].son[c] = np;

             p = e[p].f;

         }

         if(p == -) e[np].f = ;

         else{

             int q = e[p].son[c];

             if(e[p].len +  == e[q].len) e[np].f = q;

             else{

                 int nq = newnode();

                 e[nq] = e[q];

                 e[nq].len = e[p].len + ;

                 e[np].f = e[q].f = nq;

                 while(p != - && e[p].son[c] == q){

                     e[p].son[c] = nq;

                     p = e[p].f;

                 }

             }

         }

         last = np;

     }

 }sam;

 char a[maxn],b[maxn];

 ULL dp[];

 int main(){

     int kase;

     scanf("%d",&kase);

     while(kase--){

         scanf("%s%s",a,b);

         reverse(b,b + strlen(b));

         sam.init();

         node *e = sam.e;

         memset(dp,,sizeof dp);

         for(int i = ; b[i]; ++i)

             sam.add(b[i] - 'a');

         for(int i = ; i < sam.tot; ++i)

             dp[e[i].alpha] += e[i].len - e[e[i].f].len;

         sam.init();

         ULL ret = ;

         for(int i = ; a[i]; ++i)

             sam.add(a[i] - 'a');

         for(int i = ; i < sam.tot; ++i){

             ret += e[i].len - e[e[i].f].len;

             for(int j = ; j < ; ++j)

                 if(e[i].son[j] == -) ret += dp[j]*(e[i].len - e[e[i].f].len);

         }

         printf("%I64u\n",ret + );

     }

     return ;

 }

巴特西