• 【LOJ】#2525. 「HAOI2018」字串覆盖


    题解

    写后缀树真是一写就好久,然后调好久QAQ

    我们把两个串取反拼一起建后缀树,这样的话使得后缀树是正串的后缀树

    然后我们把询问挂在每个节点上,每次线段树合并,对于大于50的每次暴力跳着在线段树找,对于小于50的建出一棵树来,也就是(a[i][j])表示第(i)位往后(j)位,向下一个(a[t][j])(t >= j)连一条边
    这是一棵树,在树上倍增就行了

    说起算法来很容易吧!!!!写起来特别烦QAQ

    代码

    #include <bits/stdc++.h>
    #define fi first
    #define se second
    #define pii pair<int,int>
    #define pdi pair<db,int>
    #define mp make_pair
    #define pb push_back
    #define enter putchar('
    ')
    #define space putchar(' ')
    #define eps 1e-8
    #define MAXN 100005
    #define mo 974711
    //#define ivorysi
    using namespace std;
    typedef long long int64;
    typedef double db;
    template<class T>
    void read(T &res) {
        res = 0;char c = getchar();T f = 1;
        while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
        }
        while(c >= '0' && c <= '9') {
    	res = res * 10 + c - '0';
    	c = getchar();
        }
        res *= f;
    }
    template<class T>
    void out(T x) {
        if(x < 0) {x = -x;putchar('-');}
        if(x >= 10) {
    	out(x / 10);
        }
        putchar('0' + x % 10);
    }
    int N,LA,LB,c[MAXN * 2],Ncnt,F = 50;
    int64 K;
    char A[MAXN],B[MAXN];
    int fa[MAXN * 4][19],up[MAXN * 4][19];
    struct node {
        int cnt,len;
        node *nxt[26],*par;
    }pool[MAXN * 4],*tail = pool,*root,*last,*que[MAXN * 4];
    vector<pii > son[MAXN * 4];
    vector<int> sm[MAXN * 4];
    int ed[MAXN * 4],posA[MAXN],posB[MAXN],dis[MAXN * 4];
    int64 ans[MAXN],sum[MAXN * 4][19];
    struct qry_node {
        int s,t,len,id;
        friend bool operator < (const qry_node &a,const qry_node &b) {
    	return a.len < b.len;
        }
    };
    vector<qry_node> Qry[MAXN * 4];
    namespace segtr {
        struct node {
    	int minv,lc,rc;
        }tr[MAXN * 20];
        int Ncnt = 0,rt[MAXN * 4];
    #define LC(u) tr[u].lc
    #define RC(u) tr[u].rc
    #define MIN(u) tr[u].minv
        void update(int u) {
    	MIN(u) = LA + 1;
    	if(LC(u)) {
    	    MIN(u) = min(MIN(u),MIN(LC(u)));
    	}
    	if(RC(u)) {
    	    MIN(u) = min(MIN(u),MIN(RC(u)));
    	}
        }
        void Insert(int &u,int L,int R,int v) {
    	if(!u) u = ++Ncnt;
    	if(L == R) {MIN(u) =  v;return;}
    	int mid = (L + R) >> 1;
    	if(v <= mid) Insert(LC(u),L,mid,v);
    	else Insert(RC(u),mid + 1,R,v);
    	update(u);
        }
        int Merge(int u,int v) {
    	if(!u) return v;
    	if(!v) return u;
    	LC(u) = Merge(LC(u),LC(v));
    	RC(u) = Merge(RC(u),RC(v));
    	update(u);
    	return u;
        }
        int Query(int u,int L,int R,int p) {
    	if(!u) return LA + 1;
    	if(L >= p) return MIN(u);
    	int mid = (L + R) >> 1;
    	if(p <= mid) return min(Query(LC(u),L,mid,p),MIN(RC(u)) != 0 ? MIN(RC(u)) : LA + 1);
    	else return Query(RC(u),mid + 1,R,p);
        }
    }
    using segtr::rt;
    using segtr::Insert;
    using segtr::Merge;
    using segtr::Query;
    void build_SAM(int c,int len) {
        node *nowp = tail++,*p;
        nowp->len = len;nowp->cnt = 1;
        for(p = last ; p && !p->nxt[c] ; p = p->par) {
    	p->nxt[c] = nowp;
        }
        if(!p) nowp->par = root;
        else {
    	node *q = p->nxt[c];
    	if(q->len == p->len + 1) nowp->par = q;
    	else {
    	    node *copyq = tail++;
    	    *copyq = *q;copyq->cnt = 0;copyq->len = p->len + 1;
    	    q->par = nowp->par = copyq;
    	    for(; p && p->nxt[c] == q ; p = p->par) {
    		p->nxt[c] = copyq;
    	    }
    	}
        }
        last = nowp;
    }
    void build_SuffixTree() {
        int m = tail - pool;
        Ncnt = m;
        for(int i = 0 ; i < m ; ++i) {
    	c[pool[i].len]++;
        }
        for(int i = 1 ; i <= LA + LB ; ++i) c[i] += c[i - 1];
        for(int i = 0 ; i < m ; ++i) {
    	que[c[pool[i].len]--] = &pool[i];
        }
        for(int i = 1 ; i <= m ; ++i) {
    	int f = que[i]->par - pool + 1;
        }
        for(int i = m ; i > 1 ; --i) {
    	int u = que[i] - pool + 1;
    	int f = que[i]->par - pool + 1;
    	fa[u][0] = f;
    	son[f].pb(mp(u,que[i]->len - que[i]->par->len));
    	if(que[i]->cnt) {
    	    if(que[i]->len <= LA) {
    		ed[u] = LA - que[i]->len + 1;
    		posA[LA - que[i]->len + 1] = u;
    	    }
    	    else posB[LB - (que[i]->len - LA) + 1] = u;
    	}
        }
    }
    void dfs(int u) {
        int s = son[u].size();
        for(int i = 0 ; i < s ; ++i) {
    	dis[son[u][i].fi] = dis[u] + son[u][i].se;
    	dfs(son[u][i].fi);
        }
    }
    void Process(int u) {
        int s = son[u].size(),t;
        for(int i = 0 ; i < s ; ++i) {
    	Process(son[u][i].fi);
    	rt[u] = Merge(rt[u],rt[son[u][i].fi]);
        }
        if(ed[u]) Insert(rt[u],1,LA,ed[u]);
        sort(Qry[u].begin(),Qry[u].end());
        if(sm[u].size()) sort(sm[u].begin(),sm[u].end());
        s = Qry[u].size(),t = sm[u].size();
        for(int i = 0 ; i < s ; ++i) {
    	if((i == 0 || Qry[u][i].len != Qry[u][i - 1].len) && Qry[u][i].len <= F) {
    	    int r = 0;
    	    for(int j = 0 ; j < t ; ++j) {
    		while(r < t && sm[u][r] - sm[u][j] < Qry[u][i].len) ++r;
    		sum[sm[u][j]][0] = 0;
    		if(r >= t) up[sm[u][j]][0] = 0;
    		else {up[sm[u][j]][0] = sm[u][r];sum[sm[u][j]][0] = sm[u][r];}
    	    }
    	    for(int k = 1 ; k <= 18 ; ++k) {
    		for(int j = 0 ; j < t ; ++j) {
    		    up[sm[u][j]][k] = up[up[sm[u][j]][k - 1]][k - 1];
    		    sum[sm[u][j]][k] = sum[sm[u][j]][k - 1] + sum[up[sm[u][j]][k - 1]][k - 1];
    		}
    	    }
    	}
    	if(Qry[u][i].len <= F) {
    	    int h = Query(rt[u],1,LA,Qry[u][i].s);
    	    if(h + Qry[u][i].len - 1 <= Qry[u][i].t) {
    		ans[Qry[u][i].id] += K - h;
    		for(int k = 18 ; k >= 0 ; --k) {
    		    if(up[h][k]) {
    			if(up[h][k] + Qry[u][i].len - 1 <= Qry[u][i].t) {
    			    ans[Qry[u][i].id] += K * (1 << k) - sum[h][k];
    			    h = up[h][k];
    			}
    		    }
    		}
    	    }
    	}
    	else {
    	    int h,le = Qry[u][i].s;
    	    while(1) {
    		h = Query(rt[u],1,LA,le);
    		if(h + Qry[u][i].len - 1 <= Qry[u][i].t) {
    		    ans[Qry[u][i].id] += K - h;
    		    le = h + Qry[u][i].len;
    		}
    		else break;
    	    }
    	}
        }
    }
    void Solve() {
        read(N);read(K);
        root = last = tail++;
        scanf("%s",A + 1);scanf("%s",B + 1);
        LA = strlen(A + 1);LB = strlen(B + 1);
        int tot = 0;
        for(int i = LA ; i >= 1 ; --i) {
    	build_SAM(A[i] - 'a',++tot);
        }
        for(int i = LB ; i >= 1 ; --i) {
    	build_SAM(B[i] - 'a',++tot);
        }
        build_SuffixTree();
        for(int j = 1 ; j <= 18 ; ++j) {
    	for(int i = 1 ; i <= Ncnt ; ++i) {
    	    fa[i][j] = fa[fa[i][j - 1]][j - 1];
    	}
        }
        int s,t,l,r,q;
        dfs(1);
        read(q);
        for(int i = 1 ; i <= q ; ++i) {
    	read(s);read(t);read(l);read(r);
    	int h = posB[l];
    	for(int j = 18 ; j >= 0 ; --j) {
    	    if(dis[fa[h][j]] >= (r - l + 1)) h = fa[h][j];
    	}
    	Qry[h].pb((qry_node){s,t,r - l + 1,i});
        }
        for(int i = 1 ; i <= LA ; ++i) {
    	int t = posA[i];
    	for(int j = 18 ; j >= 0 ; --j) {
    	    if(dis[fa[t][j]] >= F) t = fa[t][j];
    	}
    	while(t) {
    	    sm[t].pb(i);
    	    t = fa[t][0];
    	}
        }
        Process(1);
        for(int i = 1 ; i <= q ; ++i) {
    	out(ans[i]);enter;
        }
    }
    int main() {
    #ifdef ivorysi
        freopen("f1.in","r",stdin);
    #endif
        Solve();
    }
    
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  • 原文地址:https://www.cnblogs.com/ivorysi/p/10049894.html
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