# 【codeforces666E】Forensic Examination 广义后缀自动机+树上倍增+线段树合并

$|S|,q\le 5\times 10^5$ ，$\sum\limits_{i=1}^m|T_i|\le 5\times 10^4$ 。

#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <algorithm>
#define N 1100000
#define lson l , mid , ls[x]
#define rson mid + 1 , r , rs[x]
using namespace std;
typedef pair<int , int> pr;
vector<int> vq[N];
int m , pos[N] , c[N][26] , dis[N] , pre[N] , tot = 1 , last = 1 , head[N] , to[N] , next[N] , cnt , fa[N][22] , deep[N] , log[N] , ls[N * 5] , rs[N * 5] , root[N] , tp , ql[N] , qr[N];
pr mx[N * 5] , ans[N];
char str[N];
void extend(int x)
{
int p = last;
if(c[p][x])
{
int q = c[p][x];
if(dis[q] == dis[p] + 1) last = q;
else
{
int nq = ++tot;
memcpy(c[nq] , c[q] , sizeof(c[q]));
dis[nq] = dis[p] + 1 , pre[nq] = pre[q] , last = pre[q] = nq;
while(p && c[p][x] == q) c[p][x] = nq , p = pre[p];
}
}
else
{
int np = last = ++tot;
dis[np] = dis[p] + 1;
while(p && !c[p][x]) c[p][x] = np , p = pre[p];
if(!p) pre[np] = 1;
else
{
int q = c[p][x];
if(dis[q] == dis[p] + 1) pre[np] = q;
else
{
int nq = ++tot;
memcpy(c[nq] , c[q] , sizeof(c[q]));
dis[nq] = dis[p] + 1 , pre[nq] = pre[q] , pre[np] = pre[q] = nq;
while(p && c[p][x] == q) c[p][x] = nq , p = pre[p];
}
}
}
}
inline void add(int x , int y)
{
to[++cnt] = y , next[cnt] = head[x] , head[x] = cnt;
}
void dfs(int x)
{
int i;
for(i = 1 ; i <= log[deep[x]] ; i ++ ) fa[x][i] = fa[fa[x][i - 1]][i - 1];
for(i = head[x] ; i ; i = next[i]) fa[to[i]][0] = x , deep[to[i]] = deep[x] + 1 , dfs(to[i]);
}
int find(int x , int d)
{
int i;
for(i = log[deep[x]] ; ~i ; i -- )
if((1 << i) <= deep[x] && dis[fa[x][i]] >= d)
x = fa[x][i];
return x;
}
inline void pushup(int x)
{
mx[x] = max(mx[ls[x]] , mx[rs[x]]);
}
void insert(int p , int l , int r , int &x)
{
if(!x) x = ++tp;
if(l == r)
{
mx[x].first ++ , mx[x].second = -p;
return;
}
int mid = (l + r) >> 1;
if(p <= mid) insert(p , lson);
else insert(p , rson);
pushup(x);
}
int merge(int l , int r , int x , int y)
{
if(!x) return y;
if(!y) return x;
if(l == r)
{
mx[x].first += mx[y].first;
return x;
}
int mid = (l + r) >> 1;
ls[x] = merge(l , mid , ls[x] , ls[y]);
rs[x] = merge(mid + 1 , r , rs[x] , rs[y]);
pushup(x);
return x;
}
pr query(int b , int e , int l , int r , int x)
{
if(b <= l && r <= e) return mx[x];
int mid = (l + r) >> 1;
if(e <= mid) return query(b , e , lson);
else if(b > mid) return query(b , e , rson);
else return max(query(b , e , lson) , query(b , e , rson));
}
void solve(int x)
{
int i;
for(i = head[x] ; i ; i = next[i]) solve(to[i]) , root[x] = merge(1 , m , root[x] , root[to[i]]);
for(i = 0 ; i < (int)vq[x].size() ; i ++ ) ans[vq[x][i]] = query(ql[vq[x][i]] , qr[vq[x][i]] , 1 , m , root[x]);
}
inline char nc()
{
static char buf[100000] , *p1 , *p2;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf , 1 , 100000 , stdin) , p1 == p2) ? EOF : *p1 ++ ;
}
{
int ret = 0; char ch = nc();
while(!isdigit(ch)) ch = nc();
while(isdigit(ch)) ret = ((ret + (ret << 2)) << 1) + (ch ^ '0') , ch = nc();
return ret;
}
inline int readstr(char *p)
{
char ch = nc() , *q = p;
while(isalpha(ch)) *q ++ = ch , ch = nc();
return q - p;
}
char pbuf[10000000] , *pp = pbuf;
inline void write(int x)
{
static int sta[20];
int top = 0;
if(!x) sta[top ++ ] = 0;
while(x) sta[top ++ ] = x % 10 , x /= 10;
while(top -- ) *pp ++ = sta[top] ^ '0';
}
int main()
{
int q , i , j , x , y;
for(i = readstr(str + 1) ; i ; i -- ) extend(str[i] - 'a') , pos[i] = last;
for(i = 1 ; i <= m ; i ++ )
{
last = 1;
for(j = readstr(str + 1) ; j ; j -- )
extend(str[j] - 'a') , insert(i , 1 , m , root[last]);
}
for(i = 2 ; i <= tot ; i ++ ) add(pre[i] , i) , log[i] = log[i >> 1] + 1;
dfs(1);
for(i = 1 ; i <= q ; i ++ )
{
ql[i] = readnum() , qr[i] = readnum() , x = readnum() , y = readnum();
vq[find(pos[x] , y - x + 1)].push_back(i);
}
solve(1);
for(i = 1 ; i <= q ; i ++ ) write(ans[i].first ? -ans[i].second : ql[i]) , *pp ++ = ' ' , write(ans[i].first) , *pp ++ = '\n';
fwrite(pbuf , 1 , pp - pbuf , stdout);
return 0;
}


|转载请注明 [原文链接][作者] ，谢谢！

posted @ 2018-04-04 20:52  GXZlegend  阅读(790)  评论(0编辑  收藏  举报