CSP14

暴力最高\(50\)吧，本地测试不太准跑得快的只得了\(10\)分，慢的却得了\(50\)分

暴力

#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
int n,a[N],m;
struct node
{
	mutable int v;
	bool operator < (const node& A)const
	{
		return v<A.v;
	}
};
multiset<node> s;
int tmp[N];
int main()
{
	speed();
	freopen("sample1.in","r",stdin);
	freopen("out.out","w",stdout);
	cin>>n>>m;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		s.insert({a[i]});
	}
	ll ans=0;
	// cout<<"*******"<<endl;
	while(s.size())
	{	
		auto it=s.end();
		int mi=1e9;
		for(int j=1;j<=m;j++)
		{
			it--;
			tmp[j]=(*it).v;
			mi=min(tmp[j],mi);
			if(it==s.begin())break;
		}
		s.erase(it,s.end());
		for(int j=1;j<=m;j++,it--)
		{
			// cout<<tmp[j]<<" ";
			tmp[j]-=mi;
			if(tmp[j]>0)s.insert({tmp[j]});
		}
		// cout<<endl;
		ans+=mi;
	}
	cout<<ans<<endl;
	return 0;
}

其实这是一道性质题，发现了很显然，\(max(\frac{\sum a_i}{m},a[n])\)
证明想象一下(LINUX)不知道怎么画图，设时间为\(x\)轴，烙饼的板为\(m\),为了尽量让每个时刻每个板上都有饼烙，所以最少要\(max(\frac{\sum a_i}{m},a[n])\)

点击查看代码

#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
ll n,a[N],m;
int main()
{
	speed();
	// freopen("sample1.in","r",stdin);
	// freopen("out.out","w",stdout);
	cin>>n>>m;ll sum=0,mx=0;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
		mx=max(mx,a[i]);
		sum+=a[i];
	}
	sort(a+1,a+1+n);
	cout<<max((ll)ceil(1.0*sum/m),mx);
	return 0;
}

T2 P9133 [THUPC 2023 初赛] 大富翁

我们可以把要求的转换为式子，设\(f(u,v)\)为\(u\)支配\(v\),题面有点歧义,不知道支配是算上过去和未来的，还是只算过去的

\[Ans=\sum_{x \in S}\sum_{y \in T}f(x,y)-\sum_{x \in S}\sum_{y \in T }f(y,x)- \sum_{x \in S}w_x\]

\[=\sum_{x \in S}(\sum_{y \in T}f(x,y)+ \sum_{y \in S}f(x,y) )-\sum_{x \in S}(\sum_{y \in T}f(y,x)+ \sum_{y \in S}f(y,x) )- \sum_{x \in S}w_x\]

\[=\sum_{x \in S}sz_x -\sum_{x \in S}dep_x -\sum_{x \in S}w_x\]

\[=\sum_{x \in S}(sz_x - dep_x - w_x) \]

点击查看代码

#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define bs bitset<70>
#define speed() ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 1e6+5;
int n,a[N],m,w[N],f[N],dep[N],sz[N];
vector <int> edge[N];
struct ca
{
	int w,id;
}q[N];
void dfs(int u)
{
	dep[u]=dep[f[u]]+1;
	sz[u]=1;
	for(auto to:edge[u])
	{
		dfs(to);
		sz[u]+=sz[to];
	}
}
bool cmp(ca a,ca b){return a.w>b.w;}
int main()
{
	speed();
	freopen("sample2.in","r",stdin);
	freopen("out.out","w",stdout);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>w[i];
	for(int i=2;i<=n;i++)cin>>f[i],edge[f[i]].pb(i);
	ll ans=0;
	dfs(1);
	for(int i=1;i<=n;i++)
	{
		q[i].id=i;q[i].w=sz[i]-dep[i]-w[i];
	}
	sort(q+1,q+1+n,cmp);
	for(int i=1;i<=n;i++)
	{
		if(i&1)
			ans+=q[i].w;
	}
	cout<<ans;
	return 0;
}