【NOI2011】阿狸的打字機——AC自動機

除做過一道dfs序維護fail樹之外，做這道題還需要想到——將詢問離線，用鄰接表儲存，即每次把所有y相同的處理。

From Hzwer：每次查詢（x,y），即爲查詢自動機上path(root->y)的所有結點中，有多少個在x的子樹中。

樹狀數組維護區間和。

反正這正解我是給跪了。bzoj Rank11。

#include <cstdio>
#include <cstring>
#include <cstring>
#include <queue>
#define N 100010
using namespace std;
char str[N];
int pos[N], lastq[N], ans[N];
int n, m;
struct QUERY {int to, next;}que[N];
inline char gc() {
    static char now[1<<16], *S, *T;
    if(S == T) {T = (S = now) + fread(now, 1, 1<<16, stdin); if(S == T) return EOF;}
    return *S++;
}
inline int read() {
    int x = 0; char c = gc();
    while(c < '0' || c > '9') c = gc();
    while(c >= '0' && c <= '9') {x = x * 10 + c - 48; c = gc();}
    return x;
}
struct ACM {
    int cnt;
    int fa[N], fail[N], next[N][26];
    queue<int> Q;
    ACM() {cnt = 0; memset(next[0], 0, sizeof(next[0]));}
    inline void build() {
        int now = 0, id = 0;
        for(int i = 1; i <= n; ++i)
            if(str[i] == 'P') pos[++id] = now;
            else if(str[i] == 'B') now = fa[now];
            else {
                if(!next[now][str[i] - 'a']) {
                    next[now][str[i] - 'a'] = ++cnt;
                    fa[cnt] = now;
                }
                now = next[now][str[i] - 'a'];
            }
    }
    inline void buildfail() {
        for(int i = 0; i < 26; ++i) if(next[0][i]) fail[next[0][i]] = 0, Q.push(next[0][i]);
        while(!Q.empty()) {
            int u = Q.front(); Q.pop();
            for(int i = 0; i < 26; ++i)
                if(next[u][i]) fail[next[u][i]] = next[fail[u]][i], Q.push(next[u][i]);
                else next[u][i] = next[fail[u]][i];
        }
    }
    struct edge {int to, next;}e[N]; int head[N], ecnt;
    inline void add(int x, int y) {e[++ecnt].to = y; e[ecnt].next = head[x]; head[x] = ecnt;}
    inline void set_tree() {for(int i = 1; i <= cnt; ++i) add(fail[i], i);}
    int st[N], ed[N], dfn;
    void dfs(int x) {
        st[x] = ++dfn;
        for(int i = head[x]; i; i = e[i].next) dfs(e[i].to);
        ed[x] = dfn;
    }
    int Bit[N];
    inline void Add_Bit(int x, int v) {for(; x <= dfn; x+= x & -x) Bit[x]+= v;}
    inline int Query_Bit(int x) {int ret = 0; for(; x; x-= x & -x) ret+= Bit[x]; return ret;}
    inline void solve() {
        int now = 0, id = 0;
        Add_Bit(st[0], 1);
        for(int i = 1; i <= n; ++i)
            if(str[i] == 'P') {
                ++id;
                for(int x = lastq[id]; x; x = que[x].next) {
                    int t = pos[que[x].to];
                    ans[x] = Query_Bit(ed[t]) - Query_Bit(st[t] - 1);
                }
            }
            else if(str[i] == 'B') Add_Bit(st[now], -1), now = fa[now];
            else now = next[now][str[i] - 'a'], Add_Bit(st[now], 1);
    }
}acm;
int main() {
    scanf("%s", str+1); n = strlen(str+1);
    acm.build(); acm.buildfail();
    m = read();
    for(int i = 1; i <= m; ++i) {
        int x = read(), y = read();
        que[i].next = lastq[y];
        lastq[y] = i;
        que[i].to = x;
    }
    acm.set_tree();
    acm.dfs(0);
    acm.solve();
    for(int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
    return 0;
}