2022年“腾讯杯”大学生程序设计竞赛 死去的 Elo 突然开始攻击我 题解
题目链接:死去的 Elo 突然开始攻击我
容易知道,如果暴力对某个区间而言进行查询,我们可以考虑使用并查集,开一个桶,每次添加一个数 \(val\),那么如果已经存在了 \(val-1\) 或者 \(val+1\) ,我们可以考虑合并 \(val\) 与 \(val-1\) 或者 \(val+1\) 在同一个集合中,查询答案就是最大的连通分量。
考虑莫队做法,如何并查集比较容易插入,但并不容易删除,但非常好撤销!所以我们考虑回滚莫队。具体原因可以详情见我的另外两篇文章,最后给出。那么就非常简单了,回滚莫队使用只加不减,在左端点回滚时撤销原来的 \(merge\) 即可。对于暴力我们可以用一个可以支持路径压缩的普通并查集即可,非暴力使用启发式合并的并查集保证复杂度。常数很小:\(O(\log{\sqrt{n}})\)。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)
#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 = 5e4 + 10;
int fa[N], siz[N], curr = 1; //需要回滚的
stack<tuple<int, int, int, int>> st; //x,y,fa[x],siz[y],preCurr
int tmpFa[N], tmpSiz[N], tmpCurr = 1; //不需要回滚的
stack<int> Cnt;
//不能路径压缩
inline int find(const int x)
{
return x == fa[x] ? x : find(fa[x]);
}
//可以路径压缩
inline int tmpFind(const int x)
{
return x == tmpFa[x] ? x : tmpFa[x] = tmpFind(tmpFa[x]);
}
//需要回滚记得用栈记录信息
inline void merge(int x, int y, const bool isNeedBack)
{
x = find(x), y = find(y);
if (siz[x] > siz[y])swap(x, y);
if (isNeedBack)st.emplace(x, y, fa[x], siz[y]);
fa[x] = y, siz[y] += siz[x], uMax(curr, siz[y]);
}
//直接合并
inline void tmpMerge(int x, int y)
{
x = tmpFind(x), y = tmpFind(y);
tmpFa[x] = y;
tmpSiz[y] += tmpSiz[x];
uMax(tmpCurr, tmpSiz[y]);
}
int pos[N]; //序列分块
//回滚莫队按照右端点升序
struct Mo
{
int l, r, id;
bool operator<(const Mo& other) const
{
return pos[l] ^ pos[other.l] ? l < other.l : r < other.r;
}
} node[N];
int cnt[N], tmpCnt[N]; //两种桶,暴力用的和非暴力的
int n, q, a[N];
int ans[N]; //答案
int e[N]; //块的右端点
//是否需要回滚的插入
inline void add(const int val, const bool isNeedBack)
{
cnt[val]++;
if (cnt[val - 1])merge(val - 1, val, isNeedBack);
if (cnt[val + 1])merge(val + 1, val, isNeedBack);
if (isNeedBack)Cnt.push(val);
}
//暴力插入
inline void tmpAdd(const int val)
{
tmpCnt[val]++;
if (tmpCnt[val - 1])tmpMerge(val - 1, val);
if (tmpCnt[val + 1])tmpMerge(val + 1, val);
}
inline void solve()
{
cin >> n >> q;
int blockSize = sqrt(n);
int blockCnt = (n + blockSize - 1) / blockSize;
forn(i, 1, n)cin >> a[i], pos[i] = (i - 1) / blockSize + 1;
forn(i, 1, n)siz[i] = tmpSiz[i] = 1, fa[i] = tmpFa[i] = i;
forn(i, 1, blockCnt)e[i] = i * blockSize;
e[blockCnt] = n;
forn(i, 1, q)
{
auto& [l,r,id] = node[i];
cin >> l >> r, id = i;
}
sortArr(node, q);
int l = 1, r = 0, last = 0;
forn(i, 1, q)
{
auto [L,R,id] = node[i];
if (pos[L] == pos[R])
{
forn(idx, L, R)tmpAdd(a[idx]);
ans[id] = tmpCurr;
tmpCurr = 1;
forn(idx, L, R)tmpSiz[a[idx]] = 1, tmpCnt[a[idx]] = 0, tmpFa[a[idx]] = a[idx]; //记得恢复
continue;
}
//和上一次左端点查询的块不同,更新新块查询
if (last != pos[L])
{
forn(idx, 1, n)siz[idx] = 1, fa[idx] = idx, cnt[idx] = 0;
curr = 1;
l = e[pos[L]] + 1;
r = l - 1;
last = pos[L];
}
while (r < R)add(a[++r], false); //右端点不需要回滚
int tmpL = l;
int preCurr = curr; //回滚答案
while (tmpL > L)add(a[--tmpL], true); //左端点需要回滚
ans[id] = curr;
//回滚信息
while (!st.empty())
{
auto [x,y,faX,sizY] = st.top();
st.pop();
fa[x] = faX, siz[y] = sizY;
}
curr = preCurr;
while (!Cnt.empty())
{
auto val = Cnt.top();
Cnt.pop();
cnt[val]--;
}
}
forn(i, 1, q)cout << ans[i] << endl;
}
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;
}
\[时间复杂度为 \ O(\log{\sqrt{n}} \times q\sqrt{n}),当然也可以用链表代替可撤销并查集优化掉常数。
\]