2024.10.19 CW 模拟赛

题目链接: https://files.cnblogs.com/files/blogs/833333/CW1019.zip?t=1729319935&download=true

T1

带权并查集模板, 考场上想到了结果写挂了qwq.

#include "iostream"
#include "cstdio"

using namespace std;

const int N=5e5+10,mod=998244353;

typedef long long ll;
typedef pair<ll,ll> pii;
#define put make_pair

template<typename T>
inline void read(T &x){
	x=0; char ch=getchar();
	while (ch>='0' and ch<='9') x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return;
}

int n,m;
ll val[N],fa[N];
ll ans=0;

pii find(int x){
	if (fa[x]==x) return put(x,val[x]);
	pii t=find(fa[x]);
	fa[x]=t.first;
	int v=val[x];
	val[x]=(val[x]+t.second-val[fa[x]]+mod)%mod;
	return put(fa[x],t.second+v);
}

inline void merge(int x,int y){
	pii f1=find(x),f2=find(y);
	(ans+=f1.second+f2.second)%=mod;
	if (f1.first!=f2.first) fa[f2.first]=f1.first;
	return;
}

int main(){
	
	read(n),read(m);
	for (int i=1;i<=n;++i) read(val[i]),fa[i]=i;
	
	for (int i=1;i<=m;++i){
		int x,y;
		read(x),read(y);
		merge(x,y);
	}
	
	printf("%lld",ans);
	
	return 0;
}

---

T2

树形 \(dp\) .

可以发现, 当最后一次跃迁以后, 不必再回到根节点.

由此, 便可定义状态转移方程:

$ f_{0/1,u} $ 表示当前在 \(u\) 点时是否选择不返回根节点. (当然, 当该点在叶子节点时最优)

\[f_{0,u}=\sum \max (0,f_{0,v}+w-k) \]

容易证明, 当跃迁的代价大于当前这条链所获得的的权值, 不走这条链是更优的.

\[f_{1,u}=\max (f_{0,v} - \max (0,f_{0,v}+w-k) + f_{1,v}) \]

#include "iostream"
#include "cstdio"
#include "vector"

using namespace std;

#define int long long 
typedef pair<int,int> pii;

template<typename T>
inline void read(T &x){
	x=0; char ch=getchar();
	while (!isdigit(ch)) ch=getchar();
	while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return;
}

const int N=1e5+10;

int n;
int k,f[2][N];
vector<pii> tr[N];

void calculate(int nw,int fa){
	for (auto to:tr[nw]){
		if (to.first==fa) continue;
		calculate(to.first,nw);
		f[0][nw]+=max(0ll,f[0][to.first]+to.second-k);
		f[1][to.first]+=to.second;
	}
	for (auto to:tr[nw]) f[1][nw]=max(f[1][nw],f[0][nw]-max(0ll,f[0][to.first]-k+to.second)+f[1][to.first]);
	return;
}

signed main(){
	
	read(n),read(k);
	for (int i=1;i<n;++i){
		int u,v,w; 
		read(u),read(v),read(w);
		tr[u].emplace_back(v,w);
		tr[v].emplace_back(u,w);
	}

	calculate(1,0);
	
	printf("%lld",f[1][1]);
	
	return 0;
}

---

T4

题解
密码: clbzdq