ABC-407-E - Most Valuable Parentheses（最有价值的括号） · znzryb

## 思路讲解

和这道题目比较像，但又不完全一样。

[CF-1011-D. Serval and Kaitenzushi Buffet](/2025/05/08/20250508-3164f2cd) 

套路那比较像，但不完全一样。

其实最重要的就是要找到一个贪心结构，我们这边发现，倒过来遍历，然后一直选左括号，一旦右括号不够了，这个时候就要选一个左括号变为右括号。又因为前面的左括号

```cpp
	ROF(i,N-1,1){
		ans+=A[i];
		pq.push(i);
		balance--;
		if(balance<0){	// 倒过来遍历，结果发现右括号数量不够了，这个时候必须牺牲一个左括号了
			ans-=A[pq.top()];
// #ifdef LOCAL
	// debug(pq.top());
    // debug(A[pq.top()]);
// #endif
			pq.pop();
			balance+=2;
		}
	}
```

## AC代码

[https://atcoder.jp/contests/abc407/submissions/66140638](https://atcoder.jp/contests/abc407/submissions/66140638)



::: 源代码

```cpp
// Problem: E - Most Valuable Parentheses
// Contest: AtCoder - AtCoder Beginner Contest 407
// URL: https://atcoder.jp/contests/abc407/tasks/abc407_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
    do { \
        std::cerr << #variable << ":"; \
        for (const auto& elem : variable) { \
            std::cerr << elem << " "; \
        } \
        std::cerr << "\n"; \
    } while (0)
#define uniVec(var) \
    do { \
        sort(var.begin(),var.end());\
        var.resize(unique(var.begin(),var.end())-var.begin());\
    } while (0)
#define debugN(var,N) \
	do{ \
		std::cerr<<#var<<":"; \
		FOR(i,1,N){ \
			std::cerr<<var[i]<<" "; \
		} \
		std::cerr<<"\n"; \
	}while(0)
#define debugMap(variable) \
    do { \
        std::cerr << #variable << ":\n"; \
        for (const auto& pair : variable) { \
            std::cerr << "  " << pair.first << " => " << pair.second << "\n"; \
        } \
        std::cerr << std::endl; \
    } while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define vecMxx(vec,x) (vec.end()-lower_bound(all(vec),x))	// vec中大于x的数有几个

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;

ll N,M,Q,X,K,lT,A[MAXN];
struct cmp{
	bool operator()(ll a,ll b){
		return A[a]>A[b];
	}
};
/*

*/
inline void solve(){
	cin>>N;
	N=2*N;
	FOR(i,1,N){
		cin>>A[i];
	}
	ll ans=0,balance=1;
	priority_queue<ll,vector<ll>,cmp> pq;
	ROF(i,N-1,1){
		ans+=A[i];
		pq.push(i);
		balance--;
		if(balance<0){	// 倒过来遍历，结果发现右括号数量不够了，这个时候必须牺牲一个左括号了
			ans-=A[pq.top()];
// #ifdef LOCAL
	// debug(pq.top());
    // debug(A[pq.top()]);
// #endif
			pq.pop();
			balance+=2;
		}
	}
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>lT;
	while(lT--){
		solve();
	}
	// solve();
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc407/submissions/66140638
*/
```


:::


## 心路历程（WA，TLE，MLE……）