P4113 [HEOI2012] 采花 题解
题目链接：采花
这题数据加强到卡了 \(2e6\) 的可持久化线段树在线做法，先给只 tle 了最后一个点的代码：
卡常参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char))return;
if (x < 0)x = -x, putchar('-');
if (x > 9)write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow)return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3() { one = tow = three = 0; }
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y)x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y)x = y;
}
namespace fasI
{
constexpr int BF_SIZE = 1 << 12;
bool IOerr = false;
inline char nc()
{
static char buf[BF_SIZE], *p1 = buf + BF_SIZE, *pend = buf + BF_SIZE;
if (p1 == pend)
{
p1 = buf;
pend = buf + fread(buf, 1, BF_SIZE,stdin);
if (pend == p1)
{
IOerr = true;
return -1;
}
}
return *p1++;
}
inline bool bla(const char ch) { return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t'; }
template <typename T>
void Rd(T& x)
{
char ch;
while (bla(ch = nc()));
T sign = 1;
if (ch == '-')sign = -1, ch = nc();
for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
x *= sign;
}
#undef BF_SIZE
}
using namespace fasI;
constexpr int N = 2e6 + 10;
struct Node
{
int left, right;
int cnt;
} node[N << 5];
#define left(x) node[x].left
#define right(x) node[x].right
#define cnt(x) node[x].cnt
int cnt;
int root[N];
int n, c, m;
inline void add(const int pre, int& curr, const int pos, const int val, const int l = 1, const int r = n)
{
node[curr = ++cnt] = node[pre];
cnt(curr) += val;
const int mid = l + r >> 1;
if (l == r)return;
if (pos <= mid)add(left(pre),left(curr), pos, val, l, mid);
else add(right(pre),right(curr), pos, val, mid + 1, r);
}
inline int query(const int L, const int R, const int pos, const int l = 1, const int r = n)
{
if (!L and !R)return 0;
if (l == r)return cnt(R) - cnt(L);
const int mid = l + r >> 1;
int rightCnt = cnt(right(R)) - cnt(right(L));
if (pos <= mid)return rightCnt + query(left(L),left(R), pos, l, mid);
return query(right(L),right(R), pos, mid + 1, r);
}
int pre[N], pre_pre[N];
int x;
inline void solve()
{
Rd(n), Rd(c), Rd(m);
forn(i, 1, n)
{
Rd(x);
root[i] = root[i - 1];
if (pre[x])add(root[i], root[i], pre[x], -1);
if (pre_pre[x])add(root[i], root[i], pre[x] = pre_pre[x], 1);
pre_pre[x] = i;
}
while (m--)
{
int l, r;
Rd(l), Rd(r);
write(endl, query(root[l - 1], root[r], l));
}
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test)solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}
这玩意挺经典的，难怪这题给离线，题目翻译下，就是问区间上有多少种数的数量 \(\ge 2\)，首先在线可持久化线段树维护一下前驱贡献类似 HH的项链。不过卡常了，因为涉及到删除，所以预处理应该是 \(2\) 倍原先建树复杂度。
讲讲离线做法，扫描线处理下查询，从左往右修改，每加入一个数，如果原先有前，我们去掉前驱的贡献，并且我们开一个桶，当然这里由于只涉及到去区间上只有两个数，其实我们再记录前驱的前驱就能表示出上一个贡献位置，如果这题改为区间内该数的数量 \(\ge k\)，可以直接开桶找前面第 \(k\) 个位置即为贡献位置。将贡献位置平移一位，用新的前第 \(2\) 个数代替原有的贡献，树状数组上最终查询即可，读入比较大，可以考虑快读或者究极快读。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-')sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char))return;
if (x < 0)x = -x, putchar('-');
if (x > 9)write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow)return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3() { one = tow = three = 0; }
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y)x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y)x = y;
}
constexpr int N = 2e6 + 10;
int bit[N];
int n, c, m;
inline void add(int x, const int val)
{
while (x <= n)bit[x] += val, x += lowBit(x);
}
inline int query(int x)
{
int ans = 0;
while (x)ans += bit[x], x -= lowBit(x);
return ans;
}
vector<pii> seg[N];
int pre[N], pre_pre[N];
int a[N];
int ans[N];
inline void solve()
{
read(n, c, m);
forn(i, 1, n)read(a[i]);
forn(i, 1, m)
{
int l, r;
read(l, r);
seg[r].emplace_back(l, i);
}
forn(r, 1, n)
{
if (pre[a[r]])add(pre[a[r]], -1);
if (pre_pre[a[r]])pre[a[r]] = pre_pre[a[r]], add(pre[a[r]], 1);
pre_pre[a[r]] = r;
for (const auto [l,id] : seg[r])ans[id] = query(r) - query(l - 1);
}
forn(i, 1, m)write(endl, ans[i]);
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test)solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}
\[时间复杂度为:\ O((n+m)\log{n})
\]
