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Hdu 5343 MZL's Circle Zhou 后缀自动机

2017-11-14 17:02 344 查看

MZL's Circle Zhou

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

Total Submission(s): 448    Accepted Submission(s): 180


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

 

Author

SXYZ
 

Source

2015 Multi-University Training Contest 5
求从两个串当中各截取一段，组成的不同子串个数。

自动机上dp可以求不同子串个数。这题为了去重，分别建两个自动机，在第一个自动机上尽量走远，若某点加上字符c不能走则加上第二个串开头为c的字符串数量。



#include <cstdio>
#include <iostream>
#include <string.h>
#include <string>
#include <map>
#include <queue>
#include <deque>
#include <vector>
#include <set>
#include <algorithm>
#include <math.h>
#include <cmath>
#include <stack>
#include <iomanip>
#define mem0(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long double ld;
typedef double db;
const int maxn=90005,maxk=26,inf=0x3f3f3f3f;
const ld pi=acos(-1.0L);
char s[maxn],t[maxn];
ull dpa[maxn*2],dpb[maxn*2];

struct SAM {
int num,last,maxlen;
struct node{
int len,fa;
int son[maxk];
} a[maxn*2];
void init() {
num=last=maxlen=0;
a[0].len=0;a[0].fa=-1;
for (int i=0;i<maxk;i++) a[0].son[i]=-1;
}
void update (int c) {
int now=++num,p;
a[now].len=a[last].len+1;
maxlen=max(maxlen,a[now].len);
memset(a[now].son,-1,sizeof(a[now].son));
for (p=last;p!=-1&&a[p].son[c]==-1;p=a[p].fa)
a[p].son[c]=now;
if (p==-1) a[now].fa=0; else {
int q=a[p].son[c];
if (a[p].len+1==a[q].len) {
a[now].fa=q;
} else {
int ne=++num;
a[ne].len=a[p].len+1;
maxlen=max(maxlen,a[ne].len);
memcpy(a[ne].son,a[q].son,sizeof(a[q].son));
a[ne].fa=a[q].fa;
for (;p!=-1&&a[p].son[c]==q;p=a[p].fa)
a[p].son[c]=ne;
a[q].fa=a[now].fa=ne;
}
}
last=now;
}
int getfa(int n) {
return a
.fa;
}
int getlen(int n) {
return a
.len;
}
int getnum() {
return num;
}
int getson(int n,int c) {
return a
.son[c];
}
};
SAM sa,sb;

bool visit[maxn*2];
void dfsb(int now) {
visit[now]=1;
dpb[now]=1;
for (int i=0;i<maxk;i++) {
int to=sb.getson(now,i);
if (to!=-1) {
if (!visit[to]) dfsb(to);
dpb[now]+=dpb[to];
}
}
}

void dfs(int now) {
visit[now]=1;
dpa[now]=1;
for (int i=0;i<maxk;i++) {
int to=sa.getson(now,i);
if (to!=-1) {
if (!visit[to]) dfs(to);
dpa[now]+=dpa[to];
} else {
int t=sb.getson(0,i);
if (t!=-1) dpa[now]+=dpb[t];
}
}
}

int main() {
int cas;
scanf("%d",&cas);
while (cas--) {
scanf("%s",s);
scanf("%s",t);
int len=strlen(t),i,j;
sa.init();sb.init();
mem0(dpa);mem0(dpb);
for (i=0;i<len;i++)
sb.update(t[i]-'a');
mem0(visit);dfsb(0);
len=strlen(s);
for (i=0;i<len;i++)
sa.update(s[i]-'a');
int now=0;
mem0(visit);
for (i=0;i<len;i++) {
now=sa.getson(now,s[i]-'a');
dpa[now]++;
}
dfs(0);
printf("%I64u\n",dpa[0]);
}
return 0;
}

                                            

                                        

                        
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