LOJ#6035. 「雅礼集训 2017 Day4」洗衣服 题解

题目链接

首先把两个过程分别贪心，然后对应匹配一下即可。

\(O(l \log n).\)

code :

#include <bits/stdc++.h>
#define LL long long
using namespace std;
template <typename T> void read(T &x){
	x = 0; char ch = getchar();
	while (!isdigit(ch)) ch = getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0',ch = getchar();
}
const int N = 1000050;
int n,m,k; LL a[N],b[N],ta[N],tb[N];
struct Node{
	int x; LL v;
	Node(int xx=0,LL vv=0){ x=xx,v=vv; }
	inline bool operator < (const Node w) const{ return v > w.v; } 
};
priority_queue<Node>H;
inline void work(int n,LL *a,LL *t){
	static LL tt[N];
	int len = 0,i; Node tmp;
	while (!H.empty()) H.pop();
	for (i = 1; i <= n; ++i) H.push(Node(i,a[i])),tt[i] = a[i];
	while (len < k){
		tmp = H.top(); H.pop(); t[++len] = tmp.v;
		i = tmp.x; tt[i] += a[i]; H.push(Node(i,tt[i])); 
	}
}
int main(){
	int i;
	read(k),read(n),read(m);
	for (i = 1; i <= n; ++i) read(a[i]);
	for (i = 1; i <= m; ++i) read(b[i]);
	work(n,a,ta); work(m,b,tb);
	LL ans = 0;
	for (i = 1; i <= k; ++i) ans = max(ans,ta[i] + tb[k+1-i]);
	cout << ans << '\n';
	return 0;
}