题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=2434
题解:
我写的是离线做法,不知道有没有在线做法。
转化一波题意,\(x\)在AC自动机上代表的字符串在\(y\)代表的字符串中出现的次数等于\(x\)在fail树的子树内有多少个点是\(y\)点Trie树上的祖先。
然后不难得到做法: 将询问离线按照\(y\)在AC自动机上的id排序(一个正常的构建Trie的方式建出来应该就满足节点的编号是它的Trie树DFS序).
在Trie树上DFS, 同时回答询问, 用树状数组维护即可
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e5;
const int S = 26;
struct Edge
{
int v,nxt;
} e[(N<<1)+3];
struct Query
{
int x,y,id;
bool operator <(const Query &arg) const
{
return y<arg.y;
}
} qr[N+3];
int fe[N+3];
int dfn[N+3];
int fa[N+3];
int son[N+3][S+3];
int fail[N+3];
char str[N+3];
int stk[N+3];
int id[N+3];
int que[N+3];
int bit[N+3];
int ord[N+3];
int ans[N+3];
int sz[N+3];
int n,q,siz,m,cnt,en;
void addval(int lrb,int val)
{
while(lrb<=cnt)
{
bit[lrb] += val;
lrb += (lrb&(-lrb));
}
}
int querysum(int rb)
{
int ret = 0;
while(rb)
{
ret += bit[rb];
rb -= (rb&(-rb));
}
return ret;
}
void addedge(int u,int v)
{
en++; e[en].v = v;
e[en].nxt = fe[u]; fe[u] = en;
}
void buildTrie()
{
int u = 0,tp = 0; stk[tp] = 0;
for(int i=1; i<=n; i++)
{
if(str[i]=='B')
{
stk[tp] = 0; tp--;
u = stk[tp];
}
else if(str[i]=='P')
{
m++; id[m] = u;
}
else
{
str[i]-=96;
if(!son[u][str[i]]) {siz++; son[u][str[i]] = siz;}
u = son[u][str[i]];
tp++; stk[tp] = u;
}
}
}
void buildACA()
{
int head = 1,tail = 0;
for(int i=1; i<=S; i++) {if(son[0][i]) {tail++; que[tail] = son[0][i];} fail[son[0][i]] = 0;}
while(head<=tail)
{
int u = que[head]; head++;
addedge(fail[u],u); addedge(u,fail[u]);
for(int i=1; i<=S; i++)
{
if(son[u][i]) {fail[son[u][i]] = son[fail[u]][i]; tail++; que[tail] = son[u][i];}
else {son[u][i] = son[fail[u]][i];}
}
}
}
void dfs(int u)
{
cnt++; dfn[u] = cnt;
sz[u] = 1;
for(int i=fe[u]; i; i=e[i].nxt)
{
if(e[i].v==fa[u]) continue;
fa[e[i].v] = u;
dfs(e[i].v);
sz[u] += sz[e[i].v];
}
}
int main()
{
scanf("%s",str+1); n = strlen(str+1);
scanf("%d",&q);
buildTrie();
buildACA();
for(int i=1; i<=q; i++)
{
int x,y; scanf("%d%d",&x,&y);
qr[i].x = id[x]; qr[i].y = id[y]; qr[i].id = i;
}
dfs(0);
sort(qr+1,qr+q+1);
int u = 0,tp = 0,j = 0; stk[tp] = 0; addval(dfn[u],0);
for(int i=1; i<=n; i++)
{
if(str[i]=='B')
{
addval(dfn[u],-1);
stk[tp] = 0; tp--;
u = stk[tp];
}
else if(str[i]=='P')
{
while(j<q && qr[j+1].y==u)
{
j++;
ans[qr[j].id] = querysum(dfn[qr[j].x]+sz[qr[j].x]-1)-querysum(dfn[qr[j].x]-1);
}
}
else
{
u = son[u][str[i]];
tp++; stk[tp] = u;
addval(dfn[u],1);
}
}
for(int i=1; i<=q; i++)
{
printf("%d\n",ans[i]);
}
return 0;
}