HDU 5343 MZL's Circle Zhou 后缀自动机+DP

Posted 2020-10-09 Meek

MZL‘s Circle Zhou

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

 

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

 

题意：

　　给你 两个字符串 A,B，你可以从A串中选一个子串或空， 从B串种选一个子串或空，组成一个新串

　　问你能组成多少不同的新串

题解：

　　麻烦的是如何去重

　　考虑一个string ， 我们只对分割点在最后位置的方案进行计数，这样即可避免重复计算

　　首先分别对A,B进行建立后缀自动机

　　预处理出对于B， 以字符X开始的不同的子串个数

　　利用A进行计数，当当前A后缀自动机状态 st  nex[st][X] = NULL 时，我们将B串所有以X开始的子串加入st状态后面，更新answer

 

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
inline long long read(){long long x=0,f=1;char ch=getchar();while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}return x*f;}
using namespace std;

const long long INF = 1e14;
const int N = 5e5+7;
const long long mod = 1000000007;

struct SAM{
    int isPlus[N * 2],endpos[N * 2];int d[N * 2];
    int cnt[N * 2];int pos[N];
    int tot,slink[2*N],trans[2*N][28],minlen[2*N],maxlen[2*N],pre;
    int newstate(int _maxlen,int _minlen,int* _trans,int _slink){
        maxlen[++tot]=_maxlen;minlen[tot]=_minlen;
        slink[tot]=_slink;
        if(_trans)for(int i=0;i<26;i++)trans[tot][i]=_trans[i];
        return tot;
    }
    int add_char(char ch,int u){
        int c=ch-‘a‘,v=u;
        int z=newstate(maxlen[u]+1,-1,NULL,0);
        isPlus[z] = 1;
        while(v&&!trans[v][c]){trans[v][c]=z;d[z]+=1;v=slink[v];}
        if(!v){ minlen[z]=1;slink[z]=1;return z;}
        int x=trans[v][c];
        if(maxlen[v]+1==maxlen[x]){slink[z]=x;minlen[z]=maxlen[x]+1;return z;}
        int y=newstate(maxlen[v]+1,-1,trans[x],slink[x]);
        slink[z]=slink[x]=y;minlen[x]=minlen[z]=maxlen[y]+1;
        while(v&&trans[v][c]==x){trans[v][c]=y;d[x]--,d[y]++;v=slink[v];}
        minlen[y]=maxlen[slink[y]]+1;
        return z;
    }
    void init_sam() {
        for(int i = 1; i <= tot; ++i)
            for(int j = 0; j < 26; ++j) trans[i][j] = 0;
        pre = tot = 1;
    }
}A,B;

unsigned long long f[N],dp[N];
char a[N],b[N];
int main() {
    int T,cas = 1;
    scanf("%d",&T);
    while(T--) {
        scanf("%s%s",a,b);
        int n = strlen(a);
        int m = strlen(b);
        A.init_sam();
        B.init_sam();
        for(int i = 0; i < n; ++i) A.pre = A.add_char(a[i],A.pre);
        for(int i = 0; i < m; ++i) B.pre = B.add_char(b[i],B.pre);

        for(int i = 0; i <= m; ++i) B.cnt[i] = 0;
        for(int i = 1; i <= B.tot; ++i) B.cnt[B.maxlen[i]] += 1;
        for(int i = 1; i <= m; ++i) B.cnt[i] += B.cnt[i-1];
        for(int i = B.tot; i >= 1; --i) B.pos[B.cnt[B.maxlen[i]]--] = i;

        for(int i = 0; i <= n; ++i) A.cnt[i] = 0;
        for(int i = 1; i <= A.tot; ++i) A.cnt[A.maxlen[i]] += 1;
        for(int i = 1; i <= n; ++i) A.cnt[i] += A.cnt[i-1];
        for(int i = A.tot; i >= 1; --i) A.pos[A.cnt[A.maxlen[i]]--] = i;
        memset(f,0,sizeof(f));
        memset(dp,0,sizeof(dp));
        for(int i = B.tot; i >= 1; --i) {
            int v = B.pos[i];
            f[v] = 1;
            for(int j = 0; j < 26; ++j) {
                if(B.trans[v][j])f[v] += f[B.trans[v][j]];
            }
        }
        for(int i = 0; i < 26; ++i)
            dp[i] = f[B.trans[1][i]];
        unsigned long long ans = 0;
        for(int i = A.tot; i >= 1; --i) {
            int v = A.pos[i];
            for(int j = 0; j < 26; ++j) {
                if(A.trans[v][j] == 0) {
                    ans += dp[j]*(A.maxlen[v] - A.minlen[v] + 1);
                }
            }
        }
        for(int i = A.tot; i >= 2; --i) ans += (A.maxlen[i] - A.minlen[i] + 1);
        cout<<ans+1<<endl;
    }
    return 0;
}