6 solutions

  • 1
    @ 2022-10-5 20:59:14
    // 思想: 先按时间升序 (金钱的顺序可升序可降序)
    // 再用 变量time 记录当前的时间,若没有项目有时间冲突,则全部游戏都可以顺利完成(废话)
    // 否则,从 i = 1 到 当前位置 枚举,看是否存在一个 项目金钱 比此项目金钱少 (可以用优先队列或小根堆优化)
    // 若存在,则选择放弃完成那个项目,再用st[N数组标记那个项目已经放弃(小根堆或优先队列则直接出队)
    // 若不存在,则放弃当前项目是最优选择
    // 最后 m - res(放弃项目的金钱) 则是答案
    #include<iostream>
    #include<algorithm>
    #include <vector>
    #include <queue>
    using namespace std;
    #define int long long
    #define x first
    #define y second
    typedef pair<int,int> PII;
    const int N = 550;
    vector<PII> v(N);
    int b[N], c[N];
    bool st[N];
    bool cmp(PII &a, PII &b)
    {
    if( a.x != b.x )	return a.x < b.x;
    return a.y > b.y;
    }
    
    signed main()
    {
    int m, n, res = 0;
    cin >> m >> n;
    for(int i = 1; i <= n; i ++ )	cin >> v[i].x;
    for(int i = 1; i <= n; i ++ )	cin >> v[i].y;
    //sort(v.begin()+1, v.begin()+1+n);
    sort(v.begin()+1, v.begin()+1+n, cmp);
    int time = 1;
    for(int i = 1; i <= n; i ++ )
    {
    if( time > v[i].x )
    {
    int idx = i;
    for(int j = 1; j < i; j ++ )
    if( v[j].y < v[idx].y && !st[j]  )
    idx = j;
    st[idx] = true;
    res += v[idx].y;
    }
    else
    time ++;
    }
    cout << m - res;
    }
    

    //小根堆版本
    #include<iostream>
    #include<algorithm>
    #include <vector>
    #include <queue>
    using namespace std;
    #define int long long
    #define x first
    #define y second
    typedef pair<int,int> PII;
    const int N = 550;
    vector<PII> v(N);
    int b[N], c[N];
    
    priority_queue<int,vector<int>,greater<int> >q;
    
    bool cmp(PII &a, PII &b)
    {
    	if( a.x != b.x )	return a.x < b.x;
    	return a.y > b.y;
    }
    
    signed main()
    {
    	int m, n, res = 0;
    	cin >> m >> n;
    	for(int i = 1; i <= n; i ++ )	cin >> v[i].x;
    	for(int i = 1; i <= n; i ++ )	cin >> v[i].y;
    	//sort(v.begin()+1, v.begin()+1+n);
    	sort(v.begin()+1, v.begin()+1+n, cmp);
    	int time = 1;
    	for(int i = 1; i <= n; i ++ )
    	{
    		if( time > v[i].x )
    		{
    			if( q.top() < v[i].y )
    			{
    				res += q.top();
    				q.pop();
    				q.push(v[i].y);
    			}
    			else
    				res += v[i].y;
    		}
    		else
    			time ++, q.push(v[i].y);
    	}
    	cout << m - res;
    }
    

    Information

    ID
    94
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    3
    Tags
    # Submissions
    95
    Accepted
    52
    Uploaded By