BZOJ 4516: [Sdoi2016]生成魔咒 [后缀自动机]

4516: [Sdoi2016]生成魔咒

题意：询问一个字符串每个前缀有多少不同的子串

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做了一下SDOI2016R1D2，题好水啊随便AK

强行开map上SAM
每个状态的贡献就是\(Max(s)-Min(s)+1\)
插入的时候维护一下就行了

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N=3e5+5, P=1e9+7;
inline int read() {
    char c=getchar(); int x=0, f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int n, s[N]; ll ans;
struct meow { map<int, int> ch; int par, val;}t[N];
int sz=1, root=1, last=1;
void extend(int c) {
    int p=last, np=++sz;
    t[np].val = t[p].val+1;
    for(; p && !t[p].ch[c]; p=t[p].par) t[p].ch[c]=np;
    if(!p) t[np].par = root;
    else {
        int q=t[p].ch[c];
        if(t[q].val == t[p].val+1) t[np].par=q;
        else {
            ans -= t[q].val - t[p].val;
            int nq=++sz; t[nq]=t[q]; t[nq].val = t[p].val+1;
            t[q].par = t[np].par = nq;
            ans += t[q].val - t[nq].val; ans ++;
            for(; p && t[p].ch[c]==q; p=t[p].par) t[p].ch[c] = nq;
        }
    }
    ans+=t[np].val - t[t[np].par].val;
    last=np;
}
int main() {
    //freopen("in","r",stdin);
    freopen("incantation.in","r",stdin);
    freopen("incantation.out","w",stdout);
    n=read();
    for(int i=1; i<=n; i++) {
        s[i]=read();
        extend(s[i]);
        printf("%lld\n",ans);
    }
}

转载于:https://www.cnblogs.com/candy99/p/6652757.html