HDU 5008 Boring String Problem ( 后缀数组 )

Posted 2022-10-10 zeronera

题意：给定一个字符串，q次询问，每次询问给定一个数k，查询原串的所有不同子串中字典序第k小的子串在原串中的开始和结束位置，若有多个答案则输出最小的开始位置，不存在输出0 0

 

后缀自动机经典问题，所以我用后缀数组

预处理sum数组记录不同字符串的个数，即sum[i] = len - sa[i] + 1 -height[i] + sum[i-1]   (len为原串长度)

对于每个k 若k > sum[n] 则输出0 0 ，即k大于不同子串的总数

否则，二分sum数组找到第k小子串所在的sa数组，即找到相应子串所位于的后缀，但是该串的开始位置不一定是最小的，所以顺着sa数组要往后找是否还有更小的答案

至于为什么只要往后，因为当前找到子串的一定是该串在sa数组中第一次，之后若还出现，height数组会将其覆盖，所以要往后找

代码很好理解，详见代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100010;
int n;
int sa[maxn], x[maxn], c[maxn], y[maxn], height[maxn];
char s[maxn];
ll sum[maxn];

void SA() //O(nlogn)

    int m = 128;
    for (int i = 0; i <= m; i++)
        c[i] = 0;
    for (int i = 1; i <= n; i++)
        c[x[i] = s[i]]++;
    for (int i = 1; i <= m; i++)
        c[i] += c[i - 1];
    for (int i = n; i >= 1; i--)
        sa[c[x[i]]--] = i;

    for (int k = 1; k <= n; k <<= 1)
    
        int p = 0;
        for (int i = 0; i <= m; i++)
            y[i] = 0;
        for (int i = n - k + 1; i <= n; i++)
            y[++p] = i;
        for (int i = 1; i <= n; i++)
            if (sa[i] > k)
                y[++p] = sa[i] - k;

        for (int i = 0; i <= m; i++)
            c[i] = 0;
        for (int i = 1; i <= n; i++)
            c[x[y[i]]]++;
        for (int i = 1; i <= m; i++)
            c[i] += c[i - 1];
        for (int i = n; i >= 1; i--)
            sa[c[x[y[i]]]--] = y[i];

        swap(x, y);
        x[sa[1]] = 1;
        p = 1;
        for (int i = 2; i <= n; ++i)
            x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k]) ? p : ++p;
        if (p >= n)
            break;
        m = p;
    


void get_height()

    int k = 0;
    //for (int i=1; i<=n; ++i) rk[sa[i]]=i; x数组即为rank数组
    for (int i = 1; i <= n; ++i)
    
        if (x[i] == 1)
            continue; //第一名height为0
        if (k)
            --k; //h[i]>=h[i-1]-1;
        int j = sa[x[i] - 1];
        while (j + k <= n && i + k <= n && s[i + k] == s[j + k])
            ++k;
        height[x[i]] = k;
    


int main()

    int q;
    while (~scanf("%s%d", s + 1, &q))
    
        n = strlen(s + 1);
        SA();
        get_height();
        height[1] = 0;
        for (int i = 1; i <= n; i++)
        
            sum[i] = n - sa[i] + 1 - height[i] + sum[i - 1];
        
        ll v, l = 0, r = 0;
        while (q--)
        
            scanf("%lld", &v);
            v = (l ^ r ^ v) + 1;
            if (v > sum[n])
                l = r = 0;
            else
            
                int pos = lower_bound(sum + 1, sum + 1 + n, v) - sum;
                v -= sum[pos - 1];
                l = sa[pos];
                r = sa[pos] + v + height[pos] - 1;
                int len = r - l + 1;
                pos++;
                while (height[pos] >= len)
                
                    if (l > sa[pos])
                    
                        l = sa[pos];
                        r = l + len - 1;
                    
                    pos++;
                
            
            printf("%lld %lld\n", l, r);
        
    
    return 0;