SPOJ SUBST1

题解参见罗穗骞的论文《后缀数组——处理字符串的有力工具》。

最近开始重拾一些重点算法，比如网络流，强连通分量，还有后缀数组（字符串和数论相关一直都是弱项）。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <numeric>

using LL = long long;

const size_t maxN = 50000 + 10;

char str[maxN];
int sa[maxN], rank[maxN], height[maxN];
size_t len;

void input()
{
    scanf("%s", str);
    len = strlen(str);
}

void buildSA()
{
    struct Node {
        int k1, k2, id;
        bool operator == (const Node& rhs) const {
            return k1 == rhs.k1 && k2 == rhs.k2;
        }
    };

    std::vector<Node> node(len), temp(len);
    std::vector<int> bucketSize;

    for (int i = 0; i < len; i++)
        rank[i] = (int)str[i];

    auto radixSort = [&] (size_t alphaSize) -> void
    {
        bucketSize.resize(alphaSize);
        std::fill(bucketSize.begin(), bucketSize.end(), 0);

        for (auto& cur: node)
            bucketSize[cur.k2] += 1;
        std::partial_sum(bucketSize.begin(), bucketSize.end(), bucketSize.begin());

        for (auto it = node.crbegin(); it != node.crend(); ++it)
            temp[--bucketSize[it->k2]] = *it;

        std::fill(bucketSize.begin(), bucketSize.end(), 0);

        for (auto& cur: temp)
            bucketSize[cur.k1] += 1;
        std::partial_sum(bucketSize.begin(), bucketSize.end(), bucketSize.begin());

        for (auto it = temp.crbegin(); it != temp.crend(); ++it)
            node[--bucketSize[it->k1]] = *it;
    };

    for (size_t step = 0; step < len; step == 0 ? step = 1 : step <<= 1)
    {
        for (int i = 0; i < len; i++)
            node[i] = {rank[i], i + step < len ? rank[i + step] : 0, i};

        radixSort(step == 0 ? size_t(128) : len);

        int last = 1;
        rank[node[0].id] = 1;
        for (size_t i = 1; i < len; i++)
        {
            if (!(node[i] == node[i - 1]))
                last += 1;
            rank[node[i].id] = last;
        }

        if (last == len)
            break;
    }

    for (int i = 0; i < len; i++)
        sa[rank[i] -= 1] = i;
}

void buildHeight()
{
    height[0] = 0;
    int last = 0;

    for (int i = 0; i < len; i++)
    {
        if (rank[i] == 0)
            continue;

        int lp = sa[rank[i] - 1], cp = i;
        int cur = std::max(0, last - 1);

        for (; str[lp + cur] == str[cp + cur]; cur++) {}
        height[rank[i]] = last = cur;
    }
}

//#include <fmt/format.h>

LL solve()
{
    buildSA();
    buildHeight();

//    for (int i = 0; i < len; i++)
//        fmt::print("sa[{i}] = {1}, rank[{i}] = {2}, height[{i}] = {3}
",
//                   fmt::arg("i", i), sa[i], rank[i], height[i]);

    LL ans = len - sa[0];
    for (int i = 1; i < len; i++)
        ans += (len - sa[i] - height[i]);

    return ans;
}

int main()
{
    int T;
    for (scanf("%d", &T); T; T--)
    {
        input();
        printf("%lld
", solve());
    }
    return 0;
}