【洛谷1117_BZOJ4650】[NOI2016] 優秀的拆分(雜湊_字尾陣列_RMQ)
題目:
分析:
定義把我校某兔姓神犇Tzz和他的妹子拆分,為“優秀的拆分”
隨便寫個雜湊就能有\(95\)分的好成績……
我的\(95\)分做法比fei較chang奇葩,不想浪費時間的可以忽略解法一qwq
解法一:
用\(n\)個vector記錄對於每個點\(i\),哪些長度\(len\)滿足\(i+2len\leq n\)且\(str[i, i+len)=str[i+len,i+2len)\)(即形如“\(AA\)”)。然後,列舉開頭\(l\)和“\(AA\)”的長度\(len\),這種情況下的的答案就是以\(l+2len\)開頭的“\(AA\)”數量(即“\(BB\)
解法二:
這題雜湊可以過的說qwq(雖然我慘遭卡常沒過去)
預處理出以每個點開始和結尾的“\(AA\)”的數量\(st[i]\)和\(ed[i]\),那麼答案就是\(\sum\limits_{i=1}^{len-1}ed[i-1]st[i]\)。
暴力列舉一個\(A\)的長度\(len\),然後每隔\(len\)個點作一個標記,共\(\lceil \frac {n}{len} \rceil\)個。可以發現任意長為\(2len\)的“\(AA\)”都會經過兩個標記。列舉標記,計算相鄰兩標記\(a\)和\(b\)開頭的字尾的\(LCP\)(最長公共字首)和兩標記結尾的字首的\(LCS\)
求\(LCP\)和\(LCS\)可以二分+雜湊解決。根據某些神奇的原理,\(O(\sum\limits _{i=1}^n \frac{n}{i})=O(n\log n)\)(貌似叫調和級數)。裡面再套個二分,複雜度\(O(n\log^2n)\)
解法三:
用字尾陣列+ST表\(O(1)\)查詢\(LCP\)和\(LCS\),複雜度\(O(n\log n)\)。(什麼,你不會後綴陣列/不會用字尾陣列查\(LCP\)和\(LCS\)?戳我:【知識總結】字尾陣列(Suffix_Array))
程式碼:
一、奇葩的雜湊\(95\)分做法
五個月前寫的程式碼,比較奇葩……
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
namespace zyt
{
typedef long long ll;
typedef pair<ll, ll> pll;
const int N = 2010, seed = 131, p[2] = {(int)1e9 + 7, (int)1e9 + 9};
ll F[N][2];
ll h[N][2];
void init()
{
F[0][0] = F[0][1] = 1;
for (int i = 1; i < N; i++)
{
F[i][0] = F[i - 1][0] * seed % p[0];
F[i][1] = F[i - 1][1] * seed % p[1];
}
}
void init(const string &str)
{
h[0][0] = h[0][1] = str[0] - 'a';
for (int t = 0; t < 2; t++)
for (int i = 1; i < str.size(); i++)
h[i][t] = (h[i - 1][t] * seed + str[i] - 'a') % p[t];
}
pll Hash(const string &str)
{
ll ans[2] = {0, 0};
for (int t = 0; t < 2; t++)
for (int i = 0; i < str.size(); i++)
ans[t] = (ans[t] * seed + str[i] - 'a') % p[t];
return make_pair(ans[0], ans[1]);
}
pll Hash(const int l, const int r)
{
if (l == 0)
return make_pair(h[r][0], h[r][1]);
else
{
int len = r - l + 1;
return make_pair(
(h[r][0] - h[l - 1][0] * F[len][0] % p[0] + p[0]) % p[0],
(h[r][1] - h[l - 1][1] * F[len][1] % p[1] + p[1]) % p[1]);
}
}
vector<int>repeat[N];
void work()
{
ios::sync_with_stdio(false);
int T;
init();
cin >> T;
while (T--)
{
int ans = 0;
string str;
cin >> str;
init(str);
for (int l = 0; l < str.size(); l++)
{
vector<int>().swap(repeat[l]);
for (int r = l; r < str.size(); r++)
{
int len = r - l + 1;
if (r + len < str.size() && Hash(l, r) == Hash(r + 1, r + len))
repeat[l].push_back(len * 2);
}
}
for (int l = 0; l < str.size(); l++)
for (int i = 0; i < repeat[l].size(); i++)
{
int r = l + repeat[l][i];
if (r < str.size())
ans += repeat[r].size();
}
cout << ans << endl;
}
}
}
int main()
{
zyt::work();
return 0;
}
二、\(O(n\log^2 n)\)的雜湊期望\(100\)實際\(95\)的做法
(我是哪裡寫掛了還是常數太大啊qwq,\(O(n\log ^2n)\)憑什麼過不了\(30000\)啊,我周圍一圈神仙都能過的qwq
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
namespace zyt
{
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int N = 3e4 + 10;
namespace Hash
{
typedef pii hash_t;
const hash_t seed = pii(29, 29), p = pii(1e9 + 7, 1e9 + 9);
pii operator + (const pii &a, const pii &b)
{
return make_pair(a.first + b.first, a.second + b.second);
}
pii operator - (const pii &a, const pii &b)
{
return make_pair(a.first - b.first, a.second - b.second);
}
pll operator * (const pii &a, const pii &b)
{
return make_pair((ll)a.first * b.first, (ll)a.second * b.second);
}
pii operator % (const pll &a, const pii &p)
{
return make_pair(a.first % p.first, a.second % p.second);
}
hash_t h[N], pow[N];
inline void init()
{
pow[0] = make_pair(1, 1);
for (int i = 1; i < N; i++)
pow[i] = pow[i - 1] * seed % p;
}
inline int ctoi(const char c)
{
return c - 'a';
}
inline hash_t ctoh(const char c)
{
return make_pair(ctoi(c), ctoi(c));
}
inline void get_hash(const string &s)
{
h[0] = ctoh(s[0]);
for (int i = 1; i < s.size(); i++)
h[i] = (h[i - 1] * seed % p + ctoh(s[i])) % p;
}
inline hash_t extract(const int l, const int r)
{
if (l == 0)
return h[r];
else
return (h[r] - h[l - 1] * pow[r - l + 1] % p + p) % p;
}
}
int st[N], ed[N];
inline int lcp(const int len, const int a, const int b)
{
using Hash::extract;
int l = 1, r = len, ans = 0;
while (l <= r)
{
int mid = (l + r) >> 1;
if (extract(a, a + mid - 1) == extract(b, b + mid - 1))
l = mid + 1, ans = mid;
else
r = mid - 1;
}
return ans;
}
inline int lcs(const int len, const int a, const int b)
{
using Hash::extract;
int l = 1, r = len, ans = 0;
while (l <= r)
{
int mid = (l + r) >> 1;
if (extract(a - mid + 1, a) == extract(b - mid + 1, b))
l = mid + 1, ans = mid;
else
r = mid - 1;
}
return ans;
}
int work()
{
ios::sync_with_stdio(false);
int T;
Hash::init();
cin >> T;
while (T--)
{
string str;
cin >> str;
memset(st, 0, sizeof(int[str.size()]));
memset(ed, 0, sizeof(int[str.size()]));
Hash::get_hash(str);
for (int i = 1; i < str.size(); i++)
for (int j = 0; j + i < str.size(); j += i)
{
int nxt = j + i;
int pre = lcp(min(i, (int)str.size() - nxt + 1), j, nxt);
int suf = lcs(min(i, j + 1), j, nxt);
int sta = min(j - suf + 1, (int)str.size() - (i << 1));
int end = min(j + pre - i, (int)str.size() - (i << 1));
if (pre + suf - 1 >= i)
{
++st[sta], --st[end + 1];
++ed[sta + (i << 1) - 1], --ed[end + (i << 1)];
}
}
for (int i = 1; i < str.size(); i++)
st[i] += st[i - 1], ed[i] += ed[i - 1];
ll ans = 0;
for (int i = 1; i < str.size(); i++)
ans += (ll)ed[i - 1] * st[i];
cout << ans << '\n';
}
return 0;
}
}
int main()
{
return zyt::work();
}
三、\(O(n\log n)\)的字尾陣列優秀做法
#include <iostream>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
namespace zyt
{
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int N = 3e4 + 10, B = 15, CH = 26;
struct ST
{
int st[N][B];
const int *arr;
static int lg2[N];
static bool lg2_built;
int min(const int a, const int b)
{
return arr[a] < arr[b] ? a : b;
}
void build(const int n, const int *_arr)
{
arr = _arr;
if (!lg2_built)
{
int tmp = 0;
for (int i = 0; i < N; i++)
{
lg2[i] = tmp;
if (i == (1 << (tmp + 1)))
++tmp;
}
lg2_built = true;
}
for (int i = n - 1; i >= 0; i--)
{
st[i][0] = i;
for (int j = 1; j <= lg2[n]; j++)
if (i + (1 << j) - 1 < n)
st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
else
break;
}
}
inline int query(const int l, const int r)
{
int len = lg2[r - l + 1];
return min(st[l][len], st[r - (1 << len) + 1][len]);
}
};
int ST::lg2[N];
bool ST::lg2_built;
struct Suffix_Array
{
int sa[N], rank[N], tp[N], count[N], height[N], kind, len;
ST st;
inline int min(const int a, const int b)
{
return height[a] < height[b] ? a : b;
}
void radix_sort()
{
static int count[N];
memset(count, 0, sizeof(int[kind]));
for (int i = 0; i < len; i++)
++count[rank[tp[i]]];
for (int i = 1; i < kind; i++)
count[i] += count[i - 1];
for (int i = len - 1; i >= 0; i--)
sa[--count[rank[tp[i]]]] = tp[i];
}
void build(const string &s)
{
len = s.size();
for (int i = 0; i < len; i++)
rank[i] = s[i] - 'a', tp[i] = i;
kind = CH;
radix_sort();
for (int tmp = 1; tmp < len; tmp <<= 1)
{
int cnt = 0;
for (int i = len - tmp; i < len; i++)
tp[cnt++] = i;
for (int i = 0; i < len; i++)
if (sa[i] >= tmp)
tp[cnt++] = sa[i] - tmp;
radix_sort();
swap(rank, tp);
rank[sa[0]] = 0;
kind = 1;
for (int i = 1; i < len; i++)
if (tp[sa[i]] == tp[sa[i - 1]] &&
(sa[i] + tmp < len && sa[i - 1] + tmp < len) &&
(tp[sa[i] + tmp] == tp[sa[i - 1] + tmp]))
rank[sa[i]] = rank[sa[i - 1]];
else
rank[sa[i]] = kind++;
if (kind == len)
break;
}
int k = 0;
for (int i = 0; i < len; i++)
{
if (!rank[i])
continue;
if (k)
--k;
int j = sa[rank[i] - 1];
while (i + k < len && j + k < len && s[i + k] == s[j + k])
++k;
height[rank[i]] = k;
}
st.build(len, height);
}
}sa1, sa2;
int st[N], ed[N];
inline int lcp(const int a, const int b)
{
int ra = sa1.rank[a], rb = sa1.rank[b];
return sa1.height[sa1.st.query(min(ra, rb) + 1, max(ra, rb))];
}
inline int lcs(const int len, const int a, const int b)
{
int ra = sa2.rank[len - a - 1], rb = sa2.rank[len - b - 1];
return sa2.height[sa2.st.query(min(ra, rb) + 1, max(ra, rb))];
}
int work()
{
ios::sync_with_stdio(false);
int T;
cin >> T;
while (T--)
{
string str, rev;
cin >> str;
memset(st, 0, sizeof(int[str.size()]));
memset(ed, 0, sizeof(int[str.size()]));
sa1.build(str);
rev = str, reverse(rev.begin(), rev.end());
sa2.build(rev);
for (int i = 1; i < str.size(); i++)
for (int j = 0; j + i < str.size(); j += i)
{
int nxt = j + i;
int pre = min(i, lcp(j, nxt));
int suf = min(i, lcs(str.size(), j, nxt));
int sta = min(j - suf + 1, (int)str.size() - (i << 1));
int end = min(j + pre - i, (int)str.size() - (i << 1));
if (pre + suf - 1 >= i)
{
++st[sta], --st[end + 1];
++ed[sta + (i << 1) - 1], --ed[end + (i << 1)];
}
}
for (int i = 1; i < str.size(); i++)
st[i] += st[i - 1], ed[i] += ed[i - 1];
ll ans = 0;
for (int i = 1; i < str.size(); i++)
ans += (ll)ed[i - 1] * st[i];
cout << ans << '\n';
}
return 0;
}
}
int main()
{
return zyt::work();
}