[TJOI2016 & HEOI2016] 字符串

[题目链接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4556

[算法]

不难发现 , 对于每个询问
        ans = max{ min{b - i + 1 , lcp(i , c) } (a <= i <= b)

不妨二分答案mid ， 那么问题就转化为求 max{ lcp(i  , c) } (a <= i <= b - mid + 1)

而我们知道 ， 所有lcp(i , j) <= k的i是连续的一段区间

可以再次通过二分求出这个区间

问题又转化为判断[a , b - mid + 1]中是否有rank值在区间[L , R]中的数

构建出后缀数组 ， 主席树维护rank值即可

时间复杂度 : O(NlogN ^ 2)

[代码]

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + ;
const int MAXLOG = ;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

#define rint register int

int n , m;
int rk[N] , rt[N] , sa[N] , cnt[N] , height[N] , lg[N];
int val[N][MAXLOG];
char s[N];

struct Presitent_Segment_Tree
{
        int sz;
        int lc[N * ] , rc[N * ] , cnt[N * ];
        Presitent_Segment_Tree()
        {
                sz = ;
        }
        inline void build(int &now , int l , int r)
        {
                now = ++sz;
                if (l == r) return;
                 int mid = (l + r) >> ;
                 build(lc[now] , l , mid);
                 build(rc[now] , mid +  , r);
        }
        inline void modify(int &now , int old , int l , int r , int x , int value)
        {
                now = ++sz;
                lc[now] = lc[old] , rc[now] = rc[old];
                cnt[now] = cnt[old] + value;
                if (l == r) return;
                int mid = (l + r) >> ;
                if (mid >= x) modify(lc[now] , lc[old] , l , mid , x , value);
                else modify(rc[now] , rc[old] , mid +  , r , x , value);
        }
        inline bool query(int rt1 , int rt2 , int l , int r , int ql , int qr)
        {
                if (ql > qr || cnt[rt1] - cnt[rt2] == )
                        return false;
                if (l == ql && r == qr)
                        return (cnt[rt1] - cnt[rt2] > );
                int mid = (l + r) >> ;
                if (mid >= qr) return query(lc[rt1] , lc[rt2] , l , mid , ql , qr);
                else if (mid +  <= ql) return query(rc[rt1] , rc[rt2] , mid +  , r , ql , qr);
                else return query(lc[rt1] , lc[rt2] , l , mid , ql , mid) | query(rc[rt1] , rc[rt2] , mid +  , r , mid +  , qr);
        }
} PST;
template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
inline void build_sa()
{
        static int x[N] , y[N];
    memset(cnt ,  , sizeof(cnt));
    for (rint i = ; i <= n; i++) ++cnt[(int)s[i]];
    for (rint i = ; i <= ; i++) cnt[i] += cnt[i - ];
    for (rint i = n; i >= ; i--) sa[cnt[(int)s[i]]--] = i;
    rk[sa[]] = ;
    for (rint i = ; i <= n; i++) rk[sa[i]] = rk[sa[i - ]] + (s[sa[i]] != s[sa[i - ]]);
    for (rint k = ; rk[sa[n]] != n; k <<= )
    {
        for (rint i = ; i <= n; i++)
            x[i] = rk[i] , y[i] = (i + k <= n) ? rk[i + k] : ;
        memset(cnt ,  , sizeof(cnt));
        for (rint i = ; i <= n; i++) ++cnt[y[i]];
        for (rint i = ; i <= n; i++) cnt[i] += cnt[i - ];
        for (rint i = n; i >= ; i--) rk[cnt[y[i]]--] = i;
        memset(cnt ,  , sizeof(cnt));
        for (rint i = ; i <= n; i++) ++cnt[x[i]];
        for (rint i = ; i <= n; i++) cnt[i] += cnt[i - ];
        for (rint i = n; i >= ; i--) sa[cnt[x[rk[i]]]--] = rk[i];
        rk[sa[]] = ;
        for (rint i = ; i <= n; i++) rk[sa[i]] = rk[sa[i - ]] + (x[sa[i]] != x[sa[i - ]] || y[sa[i]] != y[sa[i - ]]);
    }
}
inline void get_height()
{
    int k = ;
    for (rint i = ; i <= n; i++)
    {
        if (k) --k;
        int j = sa[rk[i] - ];
        while (s[i + k] == s[j + k]) ++k;
        height[rk[i]] = k;
    }
}
inline void rmq_init()
{
        for (rint i = ; i <= n; i++)
                val[i][] = height[i];
        for (rint j = ; ( << j) <= n; j++)
        {
                for (rint i = ; i + ( << j) -  <= n; i++)
                {
                        val[i][j] = min(val[i][j - ] , val[i + ( << (j - ))][j - ]);
                }
        }
}
inline int query(int x , int y)
{
        if (x > y) return ;
        int k = lg[y - x + ];
        return min(val[x][k] , val[y - ( << k) + ][k]);
}

int main()
{

        scanf("%d%d" , &n , &m);
        scanf("%s" , s + );
        build_sa();
        get_height();
        rmq_init();
        PST.build(rt[] ,  , n);
        for (rint i = ; i <= n; i++) PST.modify(rt[i] , rt[i - ] ,  , n , rk[i] , );
        for (rint i = ; i <= n; i++) lg[i] = (double)(log(i) / log(2.0));
        while (m--)
        {
                int a , b , c , d;
                read(a); read(b); read(c); read(d);
                int l =  , r = min(d - c +  , b - a + ) , ans = ;
                while (l <= r)
                {
                        int mid = (l + r) >> ;
                        int ll =  , rr = rk[c] -  , L = rk[c] , R = rk[c];
                        while (ll <= rr)
                        {
                                int md = (ll + rr) >> ;
                                if (query(md +  , rk[c]) >= mid)
                                {
                                        L = md;
                                        rr = md - ;
                                } else ll = md + ;
                        }
                        ll = rk[c] +  , rr = n , R = rk[c];
                        while (ll <= rr)
                        {
                                int md = (ll + rr) >> ;
                                if (query(rk[c] +  , md) >= mid)
                                {
                                        R = md;
                                        ll = md + ;
                                } else rr = md - ;
                        }
                        if (PST.query(rt[b - mid + ] , rt[a - ] ,  , n , L , R))
                        {
                                l = mid + ;
                                ans = mid;
                        } else r = mid - ;
                }
                printf("%d\n" , ans);
        }

        return ;

}