P3527 [POI2011] MET-Meteors 题解
题目链接：MET-Meteors
看完第一反应头大，第二反应整体二分完全能做。站在整体二分的视角思考问题，要先清楚一点，本题哪个是修改，哪个是查询。显然一开始给的 \(n\) 个数的容量，为需要查询的操作，而 \(k\) 场流星雨即为修改操作。然后第二步观察查询是否具有单调性，显然是有的，第 \(i\) 场流星雨恰好满足答案，那么 \(1 \sim i-1\) 场就不满足，\(\ge i\) 场就满足。那么这玩意就可以二分。注意到每个查询关系若干个点，可以维护一个桶记录每个查询对应的关联点。那么很显然，对于若干个查询而言，虽然单个最坏为查询 \(n\) 个点，但是均摊到每个查询上，总查询次数不超过 \(n\) 次。观察到这一点就容易了，由于只涉及到区间加，我们可以使用树状数组 \(+\) 差分区间的操作。这样一来就可以进行单点查询了。当然有些人也给出了双指针去 \(log\) 的做法，降到了单 \(\log\) 也不错，这里就使用了常见的 BIT 结合整体二分做法。稍微提下，修改和查询的顺序是不能乱排的，所以我们应该先读入了 \(n\) 个查询以后，先将后面 \(k\) 个修改进行插入整体二分列表，再插入查询。最后，二分答案给个 \([1,k+1]\)，第 \(k+1\) 场流星雨显然就是题目说的不存在。
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 = 3e6 + 10;
vector<int> pos[N];
struct Query
{
bool isQuery;
int l, r, id, val;
} q[N], qL[N], qR[N];
int ans[N];
ull bit[N];
int n, m;
inline void add(int x, const int val)
{
while (x <= m)bit[x] += val, x += lowBit(x);
}
inline ull query(int x)
{
ull ans = 0;
while (x)ans += bit[x], x -= lowBit(x);
return ans;
}
inline ull queryCnt(const int curr)
{
ull ans = 0;
for (const auto x : pos[curr])ans += query(x);
return ans;
}
inline void binary(const int L, const int R, const int idxL, const int idxR)
{
if (L == R)
{
forn(i, idxL, idxR)if (q[i].isQuery)ans[q[i].l] = L;
return;
}
const int mid = L + R >> 1;
int cntL = 0, cntR = 0;
forn(i, idxL, idxR)
{
auto& [isQuery,l,r,id,val] = q[i];
if (isQuery)
{
const ull siz = queryCnt(id);
if (val <= siz)qL[++cntL] = q[i];
else val -= siz, qR[++cntR] = q[i];
continue;
}
if (id <= mid)add(l, val), add(r + 1, -val), qL[++cntL] = q[i];
else qR[++cntR] = q[i];
}
forn(i, 1, cntL)
{
auto [isQuery,l,r,id,val] = qL[i];
if (!isQuery)add(l, -val), add(r + 1, val);
}
forn(i, 1, cntL)q[idxL + i - 1] = qL[i];
forn(i, 1, cntR)q[idxL + cntL + i - 1] = qR[i];
binary(L, mid, idxL, idxL + cntL - 1);
binary(mid + 1, R, idxL + cntL, idxR);
}
int x, k;
int a[N];
inline void solve()
{
cin >> n >> m;
forn(i, 1, m)cin >> x, pos[x].push_back(i);
int idx = 0;
forn(i, 1, n)cin >> a[i];
cin >> k;
forn(i, 1, k)
{
int l, r, val;
cin >> l >> r >> val;
if (l <= r)q[++idx] = Query(false, l, r, i, val);
else
{
q[++idx] = Query(false, l, m, i, val);
q[++idx] = Query(false, 1, r, i, val);
}
}
forn(i, 1, n)q[++idx] = Query(true, i, i, i, a[i]);
binary(1, k + 1, 1, idx);
forn(i, 1, n)
{
if (ans[i] == k + 1)cout << "NIE" << endl;
else 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;
}
\[时间复杂度为:\ O(n\log^2{n})
\]
