【题解】洛谷-3181-BZOJ-4566-[HAOI2016]找相同字符

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题目

https://www.luogu.org/problemnew/show/P3181

题意

求两个字符串相同子串的对数（位置不同看做不同）。

题解

  • 代码 1：对一个串建立 SAM，用另一个串匹配。对于每一个匹配位置，要求到根的链上的 \(EndPos\) 集合大小和。
  • 代码 2, 3：建立广义 SAM，累加每个状态匹配总数。其中代码 3 没有冗余状态。
  • 如果建立广义后缀自动机的话，有的写法会产生一些无效状态，即 len[fa[u]]==len[u] ，所以不能简单地通过基数排序来获得拓扑序了，否则会出现错误的转移。

代码1

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 2E5 + 100;
const int M = maxn << 1;

int t[M][26], len[M] = {-1}, fa[M], sz = 2, last = 1;
LL cnt[M], sum[M];
void ins(int ch) {
    int p = last, np = last = sz++;
    len[np] = len[p] + 1; cnt[np] = 1;
    for (; p && !t[p][ch]; p = fa[p]) t[p][ch] = np;
    if (!p) { fa[np] = 1; return; }
    int q = t[p][ch];
    if (len[p] + 1 == len[q]) fa[np] = q;
    else {
        int nq = sz++; len[nq] = len[p] + 1;
        memcpy(t[nq], t[q], sizeof t[0]);
        fa[nq] = fa[q];
        fa[np] = fa[q] = nq;
        for (; t[p][ch] == q; p = fa[p]) t[p][ch] = nq;
    }
}

int c[maxn] = {1}, a[M];
void rsort() {
    FOR (i, 1, sz) c[len[i]]++;
    FOR (i, 1, maxn) c[i] += c[i - 1];
    FOR (i, 1, sz) a[--c[len[i]]] = i;
}

char s[maxn];
void init() {
    FORD (i, sz - 1, 1) {
        int u = a[i];
        cnt[fa[u]] += cnt[u];
    }
    FOR (i, 2, sz) {
        int u = a[i];
        sum[u] = cnt[u] * (len[u] - len[fa[u]]) + sum[fa[u]];
    }
}
LL go() {
    LL ret = 0;
    int u = 1, l = 0;
    FOR (i, 0, strlen(s)) {
        int ch = s[i] - 'a';
        while (u && !t[u][ch]) { u = fa[u]; l = len[u]; }
        ++l; u = t[u][ch];
        if (!u) u = 1;
        ret += sum[fa[u]] + cnt[u] * (l - len[fa[u]]);
    }
    return ret;
}

int main() {
    scanf("%s", s);
    FOR (i, 0, strlen(s)) ins(s[i] - 'a');
    rsort();
    init();
    scanf("%s", s);
    cout << go() << endl;
}

代码2

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 4E5 + 100;
const int M = maxn << 1;

int t[M][26], len[M] = {-1}, fa[M], sz = 2, last = 1;
LL cnt[M][2];
void ins(int ch, int id) {
    if (t[last][ch] && len[t[last][ch]] == len[last] + 1) { last = t[last][ch]; cnt[last][id] = 1; return; }
    int p = last, np = last = sz++;
    len[np] = len[p] + 1; cnt[np][id] = 1; dbg(np, id);
    for (; p && !t[p][ch]; p = fa[p]) t[p][ch] = np;
    if (!p) { fa[np] = 1; return; }
    int q = t[p][ch];
    if (len[p] + 1 == len[q]) fa[np] = q;
    else {
        int nq = sz++; len[nq] = len[p] + 1;
        memcpy(t[nq], t[q], sizeof t[0]);
        fa[nq] = fa[q];
        fa[np] = fa[q] = nq;
        for (; t[p][ch] == q; p = fa[p]) t[p][ch] = nq;
    }
}

int c[maxn] = {1}, a[M];
void rsort() {
    FOR (i, 1, sz) c[len[i]]++;
    FOR (i, 1, maxn) c[i] += c[i - 1];
    FOR (i, 1, sz) a[--c[len[i]]] = i;
}

char s[maxn];
LL go() {
    LL ret = 0;
    FORD (i, sz - 1, 1) {
        int u = a[i];
        dbg(u, len[u], cnt[u][0], cnt[u][1], len[u] - len[fa[u]], fa[u]);
        ret += 1LL * cnt[u][0] * cnt[u][1] * (len[u] - len[fa[u]]);
        FOR (j, 0, 2) cnt[fa[u]][j] += cnt[u][j];
    }
    return ret;
}

int main() {
    FOR (i, 0, 2) {
        last = 1;
        scanf("%s", s);
        FOR (j, 0, strlen(s)) ins(s[j] - 'a', i);
    }
    rsort();
    cout << go() << endl;
}

代码3

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#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define FOR(i, x, y) for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifdef zerol
#define dbg(args...) do { cout << "\033[32;1m" << #args << " -> "; err(args); } while (0)
#else
#define dbg(...)
#endif
void err() { cout << "\033[39;0m" << endl; }
template<template<typename...> class T, typename t, typename... Args>
void err(T<t> a, Args... args) { for (auto x: a) cout << x << ' '; err(args...); }
template<typename T, typename... Args>
void err(T a, Args... args) { cout << a << ' '; err(args...); }
// -----------------------------------------------------------------------------
const int maxn = 4E5 + 100;
const int M = maxn << 1;

int t[M][26], len[M] = {-1}, fa[M], sz = 2, last = 1;
LL cnt[M][2];
void ins(int ch, int id) {
    int p = last, np = 0, nq = 0, q;
    if (!t[p][ch]) {
        np = sz++;
        len[np] = len[p] + 1;
        for (; p && !t[p][ch]; p = fa[p]) t[p][ch] = np;
    }
    if (!p) fa[np] = 1;
    else {
        q = t[p][ch];
        if (len[p] + 1 == len[q]) fa[np] = q;
        else {
            nq = sz++; len[nq] = len[p] + 1;
            memcpy(t[nq], t[q], sizeof t[0]);
            fa[nq] = fa[q];
            fa[np] = fa[q] = nq;
            for (; t[p][ch] == q; p = fa[p]) t[p][ch] = nq;
        }
    }
    last = np ? np : nq ? nq : q;
    cnt[last][id] = 1;
}

int c[maxn] = {1}, a[M];
void rsort() {
    FOR (i, 1, sz) c[len[i]]++;
    FOR (i, 1, maxn) c[i] += c[i - 1];
    FOR (i, 1, sz) a[--c[len[i]]] = i;
}

char s[maxn];
LL go() {
    LL ret = 0;
    FORD (i, sz - 1, 1) {
        int u = a[i];
        dbg(u, len[u], cnt[u][0], cnt[u][1], len[u] - len[fa[u]], fa[u]);
        ret += 1LL * cnt[u][0] * cnt[u][1] * (len[u] - len[fa[u]]);
        FOR (j, 0, 2) cnt[fa[u]][j] += cnt[u][j];
    }
    return ret;
}

int main() {
    FOR (i, 0, 2) {
        last = 1;
        scanf("%s", s);
        FOR (j, 0, strlen(s)) ins(s[j] - 'a', i);
    }
    rsort();
    cout << go() << endl;
}