Main idea of the title:
You are Robin Hood who robbed the rich to help himself. There are n rich people. The ith person has m[i] yuan of wealth. It takes p[i] yuan to buy him. You can choose from everyone. 1. Rob him and you get m[i] yuan. 2. Don't operate on him. 3. Buy him for p[i] yuan and he will absolve you of a robbery. You have a strange goal: the more people you rob, the better, but every crime you commit must be excused.
Grab k people and buy off other k people. The money you grab should be enough to buy off and maximize k.
Problem solving ideas:
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First, we found that if j < k j<k J < K and k k If k is legal, then j j j is definitely OK, so the answer is that there is a dichotomy?
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Let's take a look at the assumption that now we have chosen 2 ∗ k 2*k 2 * k people. Now we can only exchange the elements in two sets e g : steal j buy i → steal i buy j eg: steal j buy i\rightarrow steal i buy j eg: steal j and buy i → steal i and buy j
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We know S S S set is a stolen set, T T The set of T is the set bought, then S = ∑ m i , P = ∑ p j S=\sum m_i,P=\sum p_j S=∑mi,P=∑pj
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Then the legal collection is S − T ≥ 0 S-T\ge0 S−T≥0
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We just make it easier to adjust and change between segments S − T ≥ 0 S-T\ge0 S−T≥0
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exchange i and j i and j i and j → \rightarrow → ( S − m i + m j ) − ( T − p j + p i ) ≥ 0 (S-m_i+m_j)-(T-p_j+p_i)\ge0 (S−mi+mj)−(T−pj+pi)≥0
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Simplified S − T + ( m j + p j ) − ( m i + p i ) ≥ 0 S-T+(m_j+p_j)-(m_i+p_i)\ge0 S−T+(mj+pj)−(mi+pi)≥0
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According to the above formula, we must be greedy to steal m i + p i m_i+p_i mi + pi , big people are the best!!
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Now we can follow m i + p i m_i+p_i mi + pi , sort, but how k k What about k people?
That is, we know that stealing must be in the front and buying the back. Then there must be a boundary line between Zhe and zhe. Then we can enumerate and judge the boundary line. -
We from [ 1 , x ] [1,x] Steal from [1,x] [ x + 1 , n ] [x+1,n] [x+1,n] buy it inside. Just judge it
AC code
#include <bits/stdc++.h> #define mid ((l + r + 1) >> 1) #define Lson rt << 1, l , mid #define Rson rt << 1|1, mid + 1, r #define ms(a,al) memset(a,al,sizeof(a)) #define log2(a) log(a)/log(2) #define lowbit(x) ((-x) & x) #define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0) #define INF 0x3f3f3f3f #define LLF 0x3f3f3f3f3f3f3f3f #define f first #define s second #define endl '\n' using namespace std; const int N = 2e6 + 10, mod = 1e9 + 9; const int maxn = 500010; const long double eps = 1e-5; const int EPS = 500 * 500; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> PII; typedef pair<ll,ll> PLL; typedef pair<double,double> PDD; template<typename T> void read(T &x) { x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f; } template<typename T, typename... Args> void read(T &first, Args& ... args) { read(first); read(args...); } struct node { int mi, pi; bool operator < (const node & a){ if(mi + pi == a.mi + a.pi) return mi > a.mi; else return mi + pi > a.mi + a.pi; } }e[maxn]; int n; ll frontmax[maxn], bemin[maxn]; bool check(int pe) { if(!pe) return 1; ms(frontmax,0); // Steal big ms(bemin,LLF); // Buy cheap multiset<int> premax; priority_queue<int> aftmin; ll sum = 0; for(int i = 1; i <= n; ++ i) { if(premax.size() < pe) {// pe before dynamic maintenance sum += e[i].mi; premax.insert(e[i].mi); frontmax[i] = sum; continue; } if(*premax.begin() < e[i].mi) { frontmax[i] = frontmax[i-1] + e[i].mi - (*premax.begin()); premax.erase(premax.begin()); premax.insert(e[i].mi); } else frontmax[i] = frontmax[i-1]; } sum = 0; for(int i = n; i >= 1; -- i) { if(aftmin.size() < pe) { sum += e[i].pi; aftmin.push(e[i].pi); bemin[i] = sum; continue; } if(aftmin.top() > e[i].pi) { bemin[i] = bemin[i+1] + e[i].pi - aftmin.top(); aftmin.pop(); aftmin.push(e[i].pi); } else bemin[i] = bemin[i+1]; } for(int i = pe; i <= n - pe; ++ i) if(frontmax[i] >= bemin[i+1]) return 1; return 0; } int main() { IOS; cin >> n; for(int i = 1; i <= n; ++ i) cin >> e[i].mi; for(int i = 1; i <= n; ++ i) cin >> e[i].pi; sort(e+1,e+1+n); int l = 0, r = n / 2; while(l < r) { if(check(mid)) l = mid; else r = mid - 1; } cout << l; return 0; }