题解 魔法宝石

传送门

好奇怪啊，建个笛卡尔树试试？
需要枚举左右子树的点对，可以之枚举一边的？
只枚举比较小的那边？
\(O(n\log n)\)？懒，所以用 map，再加个 log 也能过

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 500010
#define fir first
#define sec second
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int a[N];

namespace force{
	int ans;
	int qmax(int l, int r) {int ans=0; for (int i=l; i<=r; ++i) ans=max(ans, a[i]); return ans;}
	void solve() {
		for (int i=1; i<=n; ++i)
			for (int j=i+1; j<=n; ++j)
				if (a[i]+a[j]==qmax(i, j)) ++ans;
		cout<<ans<<endl;
	}
}

namespace task1{
	ll ans;
	map<int, int> mp[N];
	int ls[N], rs[N], top;
	pair<int, int> sta[N];
	void dfs(int u) {
		if (ls[u]) dfs(ls[u]);
		if (rs[u]) dfs(rs[u]);
		if (mp[ls[u]].size()>mp[rs[u]].size()) swap(ls[u], rs[u]);
		for (auto it:mp[ls[u]]) if (mp[rs[u]].find(a[u]-it.fir)!=mp[rs[u]].end())
			ans+=1ll*it.sec*mp[rs[u]][a[u]-it.fir];
		for (auto it:mp[ls[u]]) mp[rs[u]][it.fir]+=it.sec;
		swap(mp[rs[u]], mp[u]);
		++mp[u][a[u]];
	}
	void solve() {
		for (int i=1; i<=n; ++i) {
			int k=top;
			pair<int, int> t={a[i], i};
			while (k && t.fir>sta[k].fir) --k;
			if (k!=top) ls[t.sec]=sta[k+1].sec;
			if (k) rs[sta[k].sec]=i;
			sta[top=++k]=t;
		}
		// cout<<"ls: "; for (int i=1; i<=n; ++i) cout<<ls[i]<<' '; cout<<endl;
		// cout<<"rs: "; for (int i=1; i<=n; ++i) cout<<rs[i]<<' '; cout<<endl;
		dfs(sta[1].sec);
		cout<<ans<<endl;
	}
}

signed main()
{
	freopen("a.in", "r", stdin);
	freopen("a.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	// force::solve();
	task1::solve();

	return 0;
}