CF526F Pudding Monsters 题解
题目链接：CF 或者 洛谷
析合树真是连续段问题的降智神器
先看下题目的一些特殊性，每行每列恰好有一个棋子。考虑特殊性，\(n \times n\) 的棋盘，那么就该判断是否有 \(n\) 个棋子，容易观察到，也就是相当于每一行并且每一列都有一个棋子。而容易知道，这些棋子所在的行或者列拿出来应当是“连续的”。
容易知道一个事实，如果你的行是连续的，并且列是连续的，那么一定能够组成答案，但如果只满足其中一个，就不一定了：
比如如图所示，这三个黑色格子是属于行连续的，但列并不连续，而它们并不能组成答案。同理，列连续但行不连续也无法组成答案。
我们得到了本题的一个贡献的充要条件，行需要连续，列也需要连续，但这个二维问题比较难以解决，我们考虑让某一维先连续，即比如以行为关键字进行排序，这样一来任意一段都是行连续了，这个时候在来考虑列连续性即可。当然了也可以这样去写：\(idx[row]=col\)，这样一来，自然按照行排好序了，因为这本质是排列，并不会出现重复数字，也是排列问题常用的一种排序方式。
接下来就转变为了一个数组上，有多少个连续段区间的问题，线段树确实处理起来比较需要思维难度，但析合树解决起来就不需要太多思考。
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如果某个点是析点，显然贡献为 \(1\)，只有它的所有儿子节点合并起来才是连续段。
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如果某个点是合点，显然贡献为 \(siz \choose 2\)，因为它的任意两个儿子节点都可以组成一个新的连续段，所以应该为任取两个儿子的方案数。
那么只需要建析合树，然后遍历一遍树节点就行了。
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 = 6e5 + 10;
constexpr int T = 25;
int idx[N];
int n;
struct
{
int MAX[N][T], MIN[N][T];
int LOG2[N];
void build()
{
forn(i, 2, n)LOG2[i] = LOG2[i >> 1] + 1;
const int k = LOG2[n] + 1;
forn(i, 1, n)MAX[i][0] = MIN[i][0] = idx[i];
forn(j, 1, k)
{
forn(i, 1, n-(1<<j)+1)
{
MAX[i][j] = max(MAX[i][j - 1], MAX[i + (1 << j - 1)][j - 1]);
MIN[i][j] = min(MIN[i][j - 1], MIN[i + (1 << j - 1)][j - 1]);
}
}
}
int query(const int l, const int r) const
{
const int k = LOG2[r - l + 1];
return max(MAX[l][k], MAX[r - (1 << k) + 1][k]) - min(MIN[l][k], MIN[r - (1 << k) + 1][k]);
}
} ST;
struct
{
struct Node
{
int Min, Add;
} node[N << 2];
#define Min(x) node[x].Min
#define Add(x) node[x].Add
void push_up(const int curr)
{
Min(curr) = min(Min(ls(curr)),Min(rs(curr)));
}
void push_down(const int curr)
{
if (Add(curr))
{
Min(ls(curr)) += Add(curr), Min(rs(curr)) += Add(curr);
Add(ls(curr)) += Add(curr), Add(rs(curr)) += Add(curr);
Add(curr) = 0;
}
}
void add(const int curr, const int l, const int r, const int val, const int s = 1, const int e = n)
{
if (l <= s and e <= r)
{
Min(curr) += val, Add(curr) += val;
return;
}
const int mid = s + e >> 1;
push_down(curr);
if (l <= mid)add(ls(curr), l, r, val, s, mid);
if (r > mid)add(rs(curr), l, r, val, mid + 1, e);
push_up(curr);
}
int query(const int curr = 1, const int l = 1, const int r = n)
{
if (l == r)return l;
const int mid = l + r >> 1;
push_down(curr);
if (!Min(ls(curr)))return query(ls(curr), l, mid);
return query(rs(curr), mid + 1, r);
}
} Seg;
struct DCT
{
int left, right, type, rightL;
DCT(const int left, const int right, const int type = 0, const int right_l = 0)
: left(left),
right(right),
type(type),
rightL(right_l)
{
}
DCT() = default;
} dct[N];
inline bool isContinue(const int l, const int r)
{
return ST.query(l, r) == r - l;
}
#define left(x) dct[x].left
#define right(x) dct[x].right
#define type(x) dct[x].type
#define rightL(x) dct[x].rightL
int cnt;
int stMax[N], stMin[N], maxCnt, minCnt;
#define maxTop idx[stMax[maxCnt]]
#define minTop idx[stMin[minCnt]]
int node[N];
vector<int> child[N];
stack<int> st;
#define root st.top()
int dctRoot;
inline void build()
{
forn(i, 1, n)
{
while (maxCnt and idx[i] >= maxTop)Seg.add(1, stMax[maxCnt - 1] + 1, stMax[maxCnt], -maxTop), --maxCnt;
while (minCnt and idx[i] <= minTop)Seg.add(1, stMin[minCnt - 1] + 1, stMin[minCnt],minTop), --minCnt;
Seg.add(1, stMax[maxCnt] + 1, i, idx[i]), Seg.add(1, stMin[minCnt] + 1, i, -idx[i]);
stMax[++maxCnt] = stMin[++minCnt] = i;
int curr;
dct[curr = node[i] = ++cnt] = DCT(i, i);
const int L = Seg.query();
while (!st.empty() and left(root) >= L)
{
if (type(root) and isContinue(rightL(root), i))
{
right(root) = i, rightL(root) = left(curr);
child[root].push_back(curr);
curr = root;
st.pop();
}
else if (isContinue(left(root), i))
{
dct[++cnt] = DCT(left(root), i, 1,left(curr));
child[cnt].push_back(root), child[cnt].push_back(curr);
curr = cnt;
st.pop();
}
else
{
child[++cnt].push_back(curr);
do child[cnt].push_back(root), st.pop();
while (!isContinue(left(root), i));
dct[cnt] = DCT(left(root), i);
child[cnt].push_back(root);
curr = cnt;
st.pop();
}
}
st.push(curr);
Seg.add(1, 1, i, -1);
}
dctRoot = root;
}
ll ans;
inline void dfs(const int curr)
{
ll siz = child[curr].size();
if (type(curr))ans += siz * (siz - 1) / 2;
else ans++;
for (const auto nxt : child[curr])dfs(nxt);
}
inline void solve()
{
cin >> n;
forn(i, 1, n)
{
int x, y;
cin >> x >> y;
idx[x] = y;
}
ST.build();
build();
dfs(dctRoot);
cout << ans;
}
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;
}
\[时间复杂度为 \ O(n\log{n})
\]
