BZOJ 3230: 相似子串

3230: 相似子串

Time Limit: 20 Sec  Memory Limit: 128 MB
Submit: 1485  Solved: 361
[Submit][Status][Discuss]

Description

 

Input

输入第1行，包含3个整数N，Q。Q代表询问组数。
第2行是字符串S。
接下来Q行，每行两个整数i和j。（1≤i≤j）。

 

Output

输出共Q行，每行一个数表示每组询问的答案。如果不存在第i个子串或第j个子串，则输出-1。

 

Sample Input

5 3
ababa
3 5
5 9
8 10

Sample Output

18
16
-1

HINT

 

样例解释

第1组询问：两个子串是“aba”,“ababa”。f = 32 + 32 = 18。

第2组询问：两个子串是“ababa”,“baba”。f = 02 + 42 = 16。

第3组询问：不存在第10个子串。输出-1。

数据范围

N≤100000，Q≤100000，字符串只由小写字母'a'~'z'组成

 

Source

后缀数组+二分+RMQ

[Submit][Status][Discuss]

 

Source里说的十分清楚了，题目本身也很水。

求出后缀数组，再把字符串reverse后求出“前缀数组”。

通过后缀数组可以对子串按排名进行定位，然后查询正反LCP即可。

 

#include <bits/stdc++.h>

template <class T>
T sqr(T x)
{
    return x*x;
}

typedef long long longint;

const int maxn = 100005;
const longint inf = 1e9;

int n, m; 
char s[maxn];
longint g[maxn];
longint pre[maxn];

class SuffixArray
{
public:
    int sa[maxn], rk[maxn], ht[maxn];

    inline void init(void)
    {
        memset(ca, 0, sizeof(ca));

        for (int i = 1; i <= n; ++i)
            ++ca[s[i]];

        for (int i = 1; i <= 300; ++i)
            ca[i] += ca[i - 1];

        for (int i = n; i >= 1; --i)
            sa[ca[s[i]]--] = i;

        rk[sa[1]] = 1;

        for (int i = 2; i <= n; ++i)
            rk[sa[i]] = rk[sa[i - 1]] + (s[sa[i]] != s[sa[i - 1]]);

        for (int l = 1; rk[sa[n]] < n; l <<= 1)
        {
            memset(ca, 0, sizeof(ca));
            memset(cb, 0, sizeof(cb));

            for (int i = 1; i <= n; ++i)
            {
                ++ca[wa[i] = rk[i]];
                ++cb[wb[i] = i + l <= n ? rk[i + l] :0];
            }
            
            for (int i = 1; i <= n; ++i)
            {
                ca[i] += ca[i - 1];
                cb[i] += cb[i - 1];
            }

            for (int i = n; i >= 1; --i)
                ta[cb[wb[i]]--] = i;

            for (int i = n; i >= 1; --i)
                sa[ca[wa[ta[i]]]--] = ta[i];

            rk[sa[1]] = 1;

            for (int i = 2; i <= n; ++i)    
                rk[sa[i]] = rk[sa[i - 1]] + (wa[sa[i]] != wa[sa[i - 1]] || wb[sa[i]] != wb[sa[i - 1]]);
        }

        for (int i = 1, j = 0; i <= n; ++i)
        {
            if (--j < 0)j = 0;
            while (s[i + j] == s[sa[rk[i] - 1] + j])++j;
            ht[rk[i]] = j;
        }
        
        build(1, 1, n);
    }

    inline int lcp(int a, int b)
    {
        a = rk[a];
        b = rk[b];

        if (a > b)
        {
            a ^= b;
            b ^= a;
            a ^= b;
        }

        return query(1, 1, n, a + 1, b);
    }
private:
    void build(int t, int l, int r)
    {
        if (l == r)
            tr[t] = ht[l];
        else
        {
            int mid = (l + r) >> 1;
            build(t << 1, l, mid);
            build(t << 1 | 1, mid + 1, r);
            tr[t] = std::min(tr[t << 1], tr[t << 1 | 1]);
        }
    }
    
    int query(int t, int l, int r, int a, int b)
    {
        if (l == a && r == b)
            return tr[t];
        else
        {
            int mid = (l + r) >> 1;
            if (b <= mid)
                return query(t << 1, l, mid, a, b);
            else if (a > mid)
                return query(t << 1 | 1, mid + 1, r, a, b);
            else
                return std::min(query(t << 1, l, mid, a, mid), query(t << 1 | 1, mid + 1, r, mid + 1, b));
        }
    }
    
    int ta[maxn], wa[maxn], wb[maxn], ca[maxn], cb[maxn], tr[maxn << 2];
}A, B;

signed main(void)
{
    scanf("%d%d%s", &n, &m, s + 1); 

    g[0] = -1;

    for (int i = 1; i <= n; ++i)
        g[i] = g[i >> 1] + 1;

    A.init();
    std::reverse(s + 1, s + 1 +n);
    B.init();

    pre[0] = 0;

    for (int i = 1; i <= n; ++i)
        pre[i] = pre[i - 1] + n - A.sa[i] + 1 - A.ht[i];
        
    for (int i = 1; i <= m; ++i)
    {
        longint lt, rt; scanf("%lld%lld", &lt, &rt);
    
        if (lt > pre[n] || rt > pre[n])
            { puts("-1"); continue; }
        
        int id1, id2, a1, b1, a2, b2;
        
        id1 = std::lower_bound(pre + 1, pre + 1 + n, lt) - pre;
        id2 = std::lower_bound(pre + 1, pre + 1 + n, rt) - pre;
        
        a1 = A.sa[id1];
        a2 = A.sa[id2];
        
        b1 = a1 + A.ht[id1] - 1 + lt - pre[id1 - 1];
        b2 = a2 + A.ht[id2] - 1 + rt - pre[id2 - 1];
        
        longint ans = 0, tmp;

        tmp = a1 == a2 ? inf : A.lcp(a1, a2);
        tmp = std::min(tmp, (longint)std::min(b1 - a1 + 1, b2 - a2 + 1));

        ans += sqr(tmp);

        tmp = b1 == b2 ? inf : B.lcp(n - b1 + 1, n - b2 + 1);
        tmp = std::min(tmp, (longint)std::min(b1 - a1 + 1, b2 - a2 + 1));

        ans += sqr(tmp);

        printf("%lld\n", ans);
    }
}

 

@Author: YouSiki