Gym

题目链接：传送门

题解：

　　利用KMP的fail失配数组，快速找到当前后缀与前缀的公共前缀点

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 2e6+10, M = 1e3+20,inf = 2e9;

int fail[N],vis[N],n,Q,cnt,op[N],QQ[N];
char a[N],b[N],ch[N][3];
int sum[N];
void build_fail() {
    memset(fail,-1,sizeof(fail));
    int j = -1;
    for(int i = 1; i < cnt; ++i) {
        while(j!=-1&&b[j+1]!=b[i]) j = fail[j];
        if(b[j+1] == b[i]) j++;
        fail[i] = j;
    }
}
int len[N];
int main() {
    scanf("%d%d",&n,&Q);
    scanf("%s",a);
    for(int i = 0; i < n; ++i) b[i] = a[i];
    cnt = n;
    len[0] = n;
    for(int i = 1; i <= Q; ++i) {
        scanf("%d",&op[i]);
        len[i] = len[i-1];
        if(op[i] == 1) {
            scanf("%s",ch[i]);
            b[cnt++] = ch[i][0];
            len[i]+=1;
        }
        else if(op[i] == 2) QQ[cnt-1] = 1;
    }
    build_fail();
    for(int i = 0; i < cnt; ++i) {
        sum[i] = QQ[i];
        if(fail[i]!=-1) {
            sum[i] += sum[fail[i]];
        }
    }
    int ccnt = n;
    for(int i = 1; i <= Q; ++i) {
        if(op[i] == 1) {
            ccnt++;
        }
        else if(op[i] == 2) {
            vis[ccnt-1] = 1;
        }
        else if(op[i] == 3) {
            int ret = 0;
            if(vis[ccnt-1]) ret = sum[ccnt-1];
            else ret = sum[ccnt-1] - QQ[ccnt-1];
            printf("%d
",ret);
        }
    }
    return 0;
}