UOJ # 395 BZOJ 5417 Luogu P4770 [NOI2018] your name (suffix automata, tree line consolidation)

NOI2019 do NOI2018 exam questions. .

Topic links: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=5417
(luogu) https://www.luogu.org/problemnew/show/P4770
(UOJ) HTTP: // uoj.ac/problem/395

Solution: actually very water, is converted into \ (S [l, r] \) and \ (T \) has a different number nature of common substrings, we are required the string \ (T \) each position \ (S [l, r] \) longest match length. Then to \ (S \) to build the SAM, the \ (T \) in \ (S \) running on, each time after a point is reached if it is not within the sub-tree \ ([l, r] \ ) between the point jump fa. Praying point (ie Right collection) can merge top-down tree line in the tree. Because they require essentially different from the number of common substrings, then the \ (T \) also established a suffix automaton, each point on statistical suffix automaton T and S is the number of common substrings. With a lot of details and hung up several times, see the code.

Time complexity \ (O (n-\ log n-) \) , \ (n-\) is the total length of the input string

String details really a lot of ah, remind yourself tomorrow if you write a string of questions must shoot! (Soon forget it, I will not likely do string)

Code

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
#define llong long long
using namespace std;

inline int read()
{
    int x=0; bool f=1; char c=getchar();
    for(;!isdigit(c);c=getchar()) if(c=='-') f=0;
    for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');
    if(f) return x;
    return -x;
}

const int N = 5e5;
const int S = 26;
const int LGN = 21;
struct SuffixAutomaton
{
    int fa[(N<<1)+3];
    int len[(N<<1)+3];
    int son[(N<<1)+3][S+1];
    int ord[(N<<1)+3];
    int buc[(N<<1)+3];
    int rtn,siz,lstpos;
    void init()
    {
        for(int i=1; i<=siz; i++)
        {
            len[i] = fa[i] = 0;
            for(int j=1; j<=S; j++) son[i][j] = 0;
        }
        rtn = siz = lstpos = 1;
    }
    int insertchar(int ch)
    {
        int p = lstpos,np; siz++; np = lstpos = siz; len[np] = len[p]+1;
        for(; p && son[p][ch]==0; p=fa[p]) {son[p][ch] = np;}
        if(p==0) {fa[np] = 1;}
        else
        {
            int q = son[p][ch];
            if(len[q]==len[p]+1) {fa[np] = q;}
            else
            {
                siz++; int nq = siz; len[nq] = len[p]+1;
                memcpy(son[nq],son[q],sizeof(son[q]));
                fa[nq] = fa[q]; fa[np] = fa[q] = nq;
                for(; p && son[p][ch]==q; p=fa[p]) {son[p][ch] = nq;}
            }
        }
        return np;
    }
    void getord()
    {
        for(int i=1; i<=siz; i++) buc[i] = 0;
        for(int i=1; i<=siz; i++) buc[len[i]]++;
        for(int i=1; i<=siz; i++) buc[i] += buc[i-1];
        for(int i=siz; i>=1; i--) ord[buc[len[i]]--] = i;
    }
} sam,samt;
struct SegmentTree
{
    int ls,rs;
} sgt[(N<<1)*LGN+3];
char a[N+3],b[N+3];
int mxl[(N<<1)+3];
int rt[(N<<1)+3];
int n,m,q,rtn,siz;

void addval(int &pos,int le,int ri,int lrb)
{
    siz++; sgt[siz] = sgt[pos]; pos = siz;
    if(le==lrb && ri==lrb) {return;}
    int mid = (le+ri)>>1;
    if(lrb<=mid) {addval(sgt[pos].ls,le,mid,lrb);}
    else {addval(sgt[pos].rs,mid+1,ri,lrb);}
}

int merge(int pos1,int pos2)
{
    if(pos1==0) return pos2; if(pos2==0) return pos1;
    siz++; int pos = siz;
    sgt[pos].ls = merge(sgt[pos1].ls,sgt[pos2].ls);
    sgt[pos].rs = merge(sgt[pos1].rs,sgt[pos2].rs);
    return pos;
}

int query(int pos,int le,int ri,int rb) //largest <=rb
{
    if(pos==0) return 0;
    if(le==ri) return le;
    int mid = (le+ri)>>1;
    if(rb<=mid) return query(sgt[pos].ls,le,mid,rb);
    else
    {
        int tmp = query(sgt[pos].rs,mid+1,ri,rb);
        return tmp ? tmp : query(sgt[pos].ls,le,mid,rb);
    }
}

int main()
{
    scanf("%s",a+1); n = strlen(a+1);
    for(int i=1; i<=n; i++) a[i] -= 96;
    sam.init();
    rtn = siz = 1;
    for(int i=1; i<=n; i++)
    {
        int u = sam.insertchar(a[i]);
        rt[u] = rtn; addval(rt[u],1,n,i);
    }
    sam.getord();
    for(int i=sam.siz; i>=1; i--)
    {
        int u = sam.ord[i];
        rt[sam.fa[u]] = merge(rt[sam.fa[u]],rt[u]);
    }
    scanf("%d",&q);
    while(q--)
    {
        for(int i=1; i<=samt.siz; i++) mxl[i] = 0;
        samt.init();
        scanf("%s",b+1); m = strlen(b+1); int lb,rb; scanf("%d%d",&lb,&rb);
        for(int i=1; i<=m; i++) b[i]-=96;
        for(int i=1; i<=m; i++) samt.insertchar(b[i]);
        samt.getord();
        int u = sam.rtn,uu = samt.rtn,cur = 0; llong ans = 0ll;
        for(int i=1; i<=m; i++)
        {
            while(u && sam.son[u][b[i]]==0) {u = sam.fa[u]; cur = sam.len[u];}
            if(u!=0) {u = sam.son[u][b[i]]; cur++;} else {u = sam.rtn; cur = 0;}
            int tmp = query(rt[u],1,n,rb);
            while(u && tmp<lb+sam.len[sam.fa[u]]) {u = sam.fa[u]; tmp = query(rt[u],1,n,rb); cur = min(sam.len[u],tmp-lb+1);}
            cur = min(cur,tmp-lb+1); //注意此处必须受到区间限制 
            while(uu && samt.son[uu][b[i]]==0) {uu = samt.fa[uu];}
            uu = uu==0 ? 1 : samt.son[uu][b[i]];
            mxl[uu] = max(mxl[uu],cur);
        }
        for(int i=samt.siz; i>=1; i--)
        {
            int uu = samt.ord[i];
            mxl[samt.fa[uu]] = max(mxl[samt.fa[uu]],mxl[uu]);
        }
        for(int i=1; i<=samt.siz; i++)
        {
            mxl[i] = max(min(mxl[i],samt.len[i]),samt.len[samt.fa[i]]); //最长匹配的长度，不得小于该点最短长度 
            int tmp = samt.len[i]-mxl[i]; //该节点上不匹配的个数
            ans += (llong)tmp;
        }
        printf("%lld\n",ans);
    }
    return 0;
}