JZOJ 3572.基因模式

首先对 \(T\) 建 SAM,对 \(1 \le i \le |S|\),求出 \(l_i = \max\{j\mid S_{i\dots j} \in {\rm substrings}(T)\}\)

\(p_i = i - l_i + 1\),显然 \(p_i\) 是单调不降的,考虑用一些整体标记来维护答案。

代码:

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#include <cstdio>
#include <cstring>
using namespace std;
const int N = 1e5;
const int S = 1 << 4;
int n,m,q;
char s[N + 5],t[N + 5],id[128],res[4];
namespace SAM
{
struct node
{
int ch[4];
int fa,len;
} sam[(N << 1) + 5];
int las = 1,tot = 1;
inline void insert(int x)
{
int cur = las,p = ++tot;
sam[p].len = sam[cur].len + 1;
for(;cur && !sam[cur].ch[x];cur = sam[cur].fa)
sam[cur].ch[x] = p;
if(!cur)
sam[p].fa = 1;
else
{
int q = sam[cur].ch[x];
if(sam[cur].len + 1 == sam[q].len)
sam[p].fa = q;
else
{
int nxt = ++tot;
sam[nxt] = sam[q],sam[nxt].len = sam[cur].len + 1,sam[p].fa = sam[q].fa = nxt;
for(;cur && sam[cur].ch[x] == q;cur = sam[cur].fa)
sam[cur].ch[x] = nxt;
}
}
las = p;
}
}
int l,cnt[S + 5],cur,tag;
long long ans;
int main()
{
id['C'] = id['c'] = 1,id['G'] = id['g'] = 2,id['T'] = id['t'] = 3;
scanf("%d%s",&q,s + 1),n = strlen(s + 1);
for(register int i = 1;i <= n;++i)
SAM::insert(id[s[i]]);
for(;q;--q)
{
l = 1,cur = tag = ans = 0,memset(cnt,0,sizeof cnt);
scanf("%s%s",s + 1,t),m = strlen(s + 1),memset(res,-1,sizeof res);
for(register char *c = t;*c;res[id[*c]] = *c >= 'A' && *c <= 'Z',++c);
for(register int i = 1,x,p = 1,len = 0;i <= m;++i)
{
x = id[s[i]];
if(SAM::sam[p].ch[x])
p = SAM::sam[p].ch[x],++len;
else
{
for(;p && !SAM::sam[p].ch[x];p = SAM::sam[p].fa);
!p ? (p = 1,len = 0) : (len = SAM::sam[p].len + 1,p = SAM::sam[p].ch[x]);
}
for(;l < i - len + 1;++l)
--cnt[cur ^ tag],cur ^= 1 << id[s[l]];
tag ^= 1 << x,cur ^= 1 << x,++cnt[tag ^ (1 << x)];
for(register int i0 = 0;i0 < 2;++i0)
if(res[0] == -1 || res[0] == i0)
for(register int i1 = 0;i1 < 2;++i1)
if(res[1] == -1 || res[1] == i1)
for(register int i2 = 0;i2 < 2;++i2)
if(res[2] == -1 || res[2] == i2)
for(register int i3 = 0;i3 < 2;++i3)
if(res[3] == -1 || res[3] == i3)
ans += cnt[(i0 | (i1 << 1) | (i2 << 2) | (i3 << 3)) ^ tag];
}
printf("%lld\n",ans);
}
}