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HDU - 5008 Boring String Problem (后缀数组+二分+RMQ)

2024-07-25 11:25:00   236人阅读

Problem Description
In this problem, you are given a string s and q queries.

For each query, you should answer that when all distinct substrings of string s were sorted lexicographically, which one is the k-th smallest. 

A substring si...j of the string s = a1a2 ...an(1 ≤ i ≤ j ≤ n) is the string aiai+1 ...aj. Two substrings sx...y and sz...w are cosidered to be distinct if sx...y ≠ Sz...w
 

Input
The input consists of multiple test cases.Please process till EOF. 

Each test case begins with a line containing a string s(|s| ≤ 105) with only lowercase letters.

Next line contains a postive integer q(1 ≤ q ≤ 105), the number of questions.

q queries are given in the next q lines. Every line contains an integer v. You should calculate the k by k = (l⊕r⊕v)+1(l, r is the output of previous question, at the beginning of each case l = r = 0, 0 < k < 263, “⊕” denotes exclusive or)
 

Output
For each test case, output consists of q lines, the i-th line contains two integers l, r which is the answer to the i-th query. (The answer l,r satisfies that sl...r is the k-th smallest and if there are several l,r available, ouput l,r which with the smallest l. If there is no l,r satisfied, output “0 0”. Note that s1...n is the whole string)
 

Sample Input
aaa
4
0
2
3
5
 

Sample Output
1 1
1 3
1 2
0 0
 

Source
2014 ACM/ICPC Asia Regional Xi‘an Online
 

题意：求第k大的子串，输出左右端点，且左端点尽量小。
思路：首先，我们可以计算出不同的子串个数，这个在论文里有的，就是
n-sa[i]-height[i]。然后我们就可以统计第i大的字符串有的子串个数，然后二分查找到第k个所在的第sa[i]后缀，接着我们可以先确定右端点的范围来RMQ查找sa[j]最小的那个，只要是满足和sa[i]后缀的lcp的长度大于len，就代表也包含这个子串了，接着就是RMQ了，坑点就是l=mid的时候的多一个判断

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
//typedef long long ll;
typedef __int64 ll;
using namespace std;
const int maxn = 100010;

int sa[maxn]; 
int t1[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];

void build_sa(int s[], int n, int m) {
    int i, j, p, *x = t1, *y = t2;
    for (i = 0; i < m; i++) c[i] = 0;
    for (i = 0; i < n; i++) c[x[i] = s[i]]++;
    for (i = 1; i < m; i++) c[i] += c[i-1];
    for (i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;

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

        swap(x, y);
        p = 1, x[sa[0]] = 0;
        for (i = 1; i < n; i++) 
            x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;

        if (p >= n) break;
        m = p;
    }
}

void getHeight(int s[],int n) {
    int i, j, k = 0;
    for (i = 0; i <= n; i++)
        rank[sa[i]] = i;

    for (i = 0; i < n; i++) {
        if (k) k--;
        j = sa[rank[i]-1];
        while (s[i+k] == s[j+k]) k++;
        height[rank[i]] = k;
    }
}
int dp[maxn][30];
char str[maxn];
int r[maxn], ind[maxn][30];
ll b[maxn];

void initRMQ(int n) {
    int m = floor(log(n+0.0) / log(2.0));  
    for (int i = 1; i <= n; i++) 
        dp[i][0] = height[i];  

    for (int i = 1; i <= m; i++) {  
        for (int j = n; j; j--) {  
            dp[j][i] = dp[j][i-1];  
            if (j+(1<<(i-1)) <= n)  
                dp[j][i] = min(dp[j][i], dp[j+(1<<(i-1))][i-1]);  
        }  
    }  
}

int lcp(int l, int r) {
    int a = rank[l], b = rank[r];  
    if (a > b) 
        swap(a,b);  
    a++;  
    int m = floor(log(b-a+1.0) / log(2.0));  
    return min(dp[a][m], dp[b-(1<<m)+1][m]);  
}

void init(int n) {
    int m = floor(log(n+0.0) / log(2.0));  
    for (int i = 1; i <= n; i++) 
        ind[i][0] = sa[i];  

    for (int i = 1; i <= m; i++) {  
        for (int j = n; j; j--) {  
            ind[j][i] = ind[j][i-1];  
            if (j+(1<<(i-1)) <= n)  
                ind[j][i] = min(ind[j][i], ind[j+(1<<(i-1))][i-1]);  
        }  
    }  
}

int rmq(int a, int b) {
    int m = floor(log(b-a+1.0) / log(2.0));  
    return min(ind[a][m], ind[b-(1<<m)+1][m]);  
}

int main() {
    while (scanf("%s", str) != EOF) {
        int n = strlen(str);    
        for (int i = 0; i <= n; i++)
            r[i] = str[i];
        build_sa(r, n+1, 128);
        getHeight(r, n);
        initRMQ(n);
        init(n);

        b[0] = 0;
        for (int i = 1; i <= n; i++)
            b[i] = b[i-1] + n - sa[i] - height[i];

        int m;
        scanf("%d", &m);
        ll k;
        int lastl = 0, lastr = 0;
        while (m--) {
            scanf("%I64d", &k);
            k = (k ^ lastl ^ lastr)  + 1;
            if (k > b[n]) {
                printf("0 0\n");
                lastl = 0;
                lastr = 0;
                continue;
            }
            int id = lower_bound(b+1, b+1+n, k) - b;
            k -= b[id-1];
            int len = height[id] + k;
            int ll = id;
            int rr = id;
            int L = id, R = n;
            while (L <= R) {
                int mid = (L + R) / 2;
                if (sa[id] == sa[mid] || lcp(sa[id], sa[mid]) >= len) {
                    rr = mid;
                    L = mid + 1;
                }
                else R = mid - 1;
            }

            int ansl = rmq(ll, rr) + 1;
            int ansr = ansl + len - 1;
            printf("%d %d\n", ansl, ansr);
            lastl = ansl;
            lastr = ansr;
        }
    }    
    return 0;
}




HDU - 5008 Boring String Problem (后缀数组+二分+RMQ)


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