4556: [Tjoi2016&Heoi2016]字符串

4556: [Tjoi2016&Heoi2016]字符串

链接

分析：

　　首先可以二分这个长度。此时需要判断是否存在一个以b结尾的前缀，满足与[c,d]的lcp大于等于mid。

　　如果我们把串翻转，那么就是判断是否存在一个以b开始的后缀，这样可以建出SAM，线段树维护每个点的right集合。此时在从包含[1,b]这个状态的点沿着parent树往上跳，那么长度会减少，相应的right集合会变大，于是可以跳到第一个长度满足的点，right集合也是满足的点中最大的，看这个点中是否存在[a,b-mid+1]的位置。

　　SA的做法：求出height数组，二分一个长度mid，找到[c,d]的rnk设为p，从p往左往右找满足lcp长度都大于等于mid的区间[l,r]，然后判断[l,r]中是否在[a,b-mid+1]中出现过。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
}

const int N = ;

int ls[N * ], rs[N * ], Root[N], pos[N], ch[N][], len[N], fa[N], tmp[N], xl[N], f[N][];
int TreeIndex, Index = , Last = , n;
char s[N];

void Insert(int l,int r,int &now,int p) {
    now = ++TreeIndex;
    if (l == r) return ;
    int mid = (l + r) >> ;
    if (p <= mid) Insert(l, mid, ls[now], p);
    else Insert(mid + , r, rs[now], p);
}
int query(int l,int r,int now,int L,int R) {
    if (!now) return ;
    if (L <= l && r <= R) return ;
    int mid = (l + r) >> ;
    if (L <= mid && query(l, mid, ls[now], L, R)) return ;
    if (R > mid && query(mid + , r, rs[now], L, R)) return ;
    return ;
}
int Merge(int x,int y) {
    if (!x || !y) return x + y;
    int z = ++TreeIndex;
    ls[z] = Merge(ls[x], ls[y]);
    rs[z] = Merge(rs[x], rs[y]);
    return z;
}
void extend(int c,int i) {
    int np = ++Index, p = Last; pos[i] = np; // pos[i] = np; !!!!!
    len[np] = len[p] + ;
    for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np;
    if (!p) fa[np] = ;
    else {
        int q = ch[p][c];
        if (len[q] == len[p] + ) fa[np] = q;
        else {
            int nq = ++Index;
            memcpy(ch[nq], ch[q], sizeof(ch[q]));
            fa[nq] = fa[q];
            fa[np] = fa[q] = nq;
            len[nq] = len[p] + ;
            for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq;
        }
    }
    Last = np;
    Insert(, n, Root[np], i);
}
bool check(int mid,int x,int l,int r) {
    for (int i = ; ~i; --i) if (len[f[x][i]] >= mid) x = f[x][i];
    return query(, n, Root[x], l, r);
}
int main() {
    n = read();int m = read();
    scanf("%s", s + );
    reverse(s + , s + n + );
    for (int i = ; i <= n; ++i) extend(s[i] - 'a', i);
    for (int i = ; i <= Index; ++i) tmp[len[i]] ++;
    for (int i = ; i <= n; ++i) tmp[i] += tmp[i - ];
    for (int i = Index; i >= ; --i) xl[tmp[len[i]]--] = i;
    for (int i = Index; i > ; --i) {
        int x = xl[i];
        Root[fa[x]] = Merge(Root[fa[x]], Root[x]);
    }
    for (int i = ; i <= Index; ++i) f[i][] = fa[i];
    for (int j = ; j <= ; ++j)
        for (int i = ; i <= Index; ++i) f[i][j] = f[f[i][j - ]][j - ];
    while (m --) {
        int a = n - read() + , b = n - read() + , c = n - read() + , d = n - read() + ;
        swap(a, b); swap(c, d);
        int l = , r = min(d - c + , b - a + ), ans = ;
        while (l <= r) {
            int mid = (l + r) >> ;
            if (check(mid, pos[d], a + mid - , b)) ans = mid, l = mid + ;
            else r = mid - ;
        }
        printf("%d\n", ans);
    }
    return ;
}