C++——计算不同的非空子串个数

计算不同的非空子串

计算方法

这道题是我在BCSP-X小高组的题目中发现的一道

没事闲的就写了代码和思路：

代码

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>

using namespace std;

// 用于存储后缀数组的比较函数
struct Suffix {
    int index;
    int rank[2];
};

// 比较函数
int cmp(struct Suffix a, struct Suffix b) {
    if (a.rank[0] == b.rank[0])
        return a.rank[1] < b.rank[1];
    else
        return a.rank[0] < b.rank[0];
}

// 构建后缀数组
vector<int> buildSuffixArray(string s) {
    int n = s.size();
    struct Suffix suffixes[n];

    for (int i = 0; i < n; i++) {
        suffixes[i].index = i;
        suffixes[i].rank[0] = s[i] - 'a';
        suffixes[i].rank[1] = (i + 1 < n) ? (s[i + 1] - 'a') : -1;
    }

    sort(suffixes, suffixes + n, cmp);

    int ind[n];

    for (int k = 4; k < 2 * n; k = k * 2) {
        int rank = 0;
        int prev_rank = suffixes[0].rank[0];
        suffixes[0].rank[0] = rank;
        ind[suffixes[0].index] = 0;

        for (int i = 1; i < n; i++) {
            if (suffixes[i].rank[0] == prev_rank && suffixes[i].rank[1] == suffixes[i - 1].rank[1]) {
                prev_rank = suffixes[i].rank[0];
                suffixes[i].rank[0] = rank;
            } else {
                prev_rank = suffixes[i].rank[0];
                suffixes[i].rank[0] = ++rank;
            }
            ind[suffixes[i].index] = i;
        }

        for (int i = 0; i < n; i++) {
            int nextIndex = suffixes[i].index + k / 2;
            suffixes[i].rank[1] = (nextIndex < n) ? suffixes[ind[nextIndex]].rank[0] : -1;
        }

        sort(suffixes, suffixes + n, cmp);
    }

    vector<int> suffixArr(n);
    for (int i = 0; i < n; i++)
        suffixArr[i] = suffixes[i].index;

    return suffixArr;
}

// 构建LCP数组
vector<int> buildLCPArray(string s, vector<int> suffixArr) {
    int n = s.size();
    vector<int> lcp(n, 0);
    vector<int> invSuffix(n, 0);

    for (int i = 0; i < n; i++)
        invSuffix[suffixArr[i]] = i;

    int k = 0;

    for (int i = 0; i < n; i++) {
        if (invSuffix[i] == n - 1) {
            k = 0;
            continue;
        }

        int j = suffixArr[invSuffix[i] + 1];

        while (i + k < n && j + k < n && s[i + k] == s[j + k])
            k++;

        lcp[invSuffix[i]] = k;

        if (k > 0)
            k--;
    }

    return lcp;
}

// 计算不同非空子串的数量
int countUniqueSubstrings(string s) {
    int n = s.size();
    vector<int> suffixArr = buildSuffixArray(s);
    vector<int> lcp = buildLCPArray(s, suffixArr);

    int totalSubstrings = n * (n + 1) / 2;
    int lcpSum = 0;

    for (int i = 0; i < n; i++)
        lcpSum += lcp[i];

    return totalSubstrings - lcpSum;
}

// 主函数
int main() {
    string s;
    cin >> s;
    cout << "不同非空子串的数量: " << countUniqueSubstrings(s) << endl;
    return 0;
}

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