换根dp 2021江西 gym103366I. Homework

最新推荐文章于 2023-08-04 21:42:11 发布

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最新推荐文章于 2023-08-04 21:42:11 发布

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这篇博客探讨了两个与树形结构相关的算法问题。第一个问题是关于求解在一个由正整数构成的序列中，对于连续子序列，找到能最大化分割数的最大质因数。第二个问题是寻找一棵树的根节点，使得以该节点为根时，所有节点的深度之和最大。这两个问题都涉及到深度优先搜索（DFS）的应用，通过递归地遍历树的节点来计算和优化解决方案。

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I. Homework

Tsiying has a sequence of positive integers with a length of n and quickly calculates all the factors of each number. In order to exercise his factor calculation ability, he has selected q consecutive subsequences from the sequence and found a positive integer p greater than 1 for each subsequence, so that p can divide as many numbers in this subsequence as possible. He has also found that p may have more than one.
So the question is, how many numbers in each subsequence can be divided at most?
Input
The first line contains an integer T (1≤T≤5×104), indicating that there is T test cases next.
The first line of each test cases has two positive integers n (1≤n≤5×104), q (1≤q≤5×104).
Next line n integers ai (1≤i≤n,1≤ai≤1×106), which representing the numbers in this sequence. The two adjacent numbers are separated by a space.
Each of the next q lines contains two integers l,r (1≤l≤r≤n), representing a subsequence being queried, al,al+1,⋯,ar, and l,r are separated by a space.
The input guarantees that the sum of n does not exceed 5×104 and the sum of q does not exceed 5×104.
Output
For each test case, output q lines, each line contains a positive integer, indicating the answer.

有n个人组成有n-1条边的无根树，点权代表自己做作业的时间，边权代表去抄作业需要花的时间。
问让所有人都做完作业需要花费的时间。
$f [u] = m a x (w [u], f [v] + d (u, v))$
换根dp。

#include<bits/stdc++.h>
using namespace std;

#define read(x) scanf("%d",&x)
#define maxn int(1e5)
#define ll long long

struct Edge{
	int y,z;
	Edge(){}
	Edge(int _y,int _z) {y=_y,z=_z;}
};

struct ID{
	int x,y,p1,p2;
	ID() {}
	ID(int _x,int _y,int _p1,int _p2) {x=_x,y=_y,p1=_p1,p2=_p2;} 
};

int n,q;
int w[maxn+5];
vector<Edge> a[maxn+5];
ID id[maxn+5];	//第i条边在vector中的位置 

ll f[maxn+5];	//完成作业的最小时间 

void dfs1(int x,int fa) {
	f[x]=w[x];	//自己做作业
	for(int i=0;i<a[x].size();i++) {
		int y=a[x][i].y,z=a[x][i].z;
		if(y==fa) continue;
		dfs1(y,x);
		f[x]=min(f[x],f[y]+z);
	} 
}

void dfs2(int x,int fa) {
	for(int i=0;i<a[x].size();i++) {
		int y=a[x][i].y,z=a[x][i].z;
		if(y==fa) continue;
		f[y]=min(f[y],f[x]+z);
		dfs2(y,x);
	}
}

int main() {
	read(n),read(q);
	for(int i=1;i<=n;i++) read(w[i]);
	for(int i=1;i<n;i++) {
		int x,y,z;
		read(x),read(y),read(z);
		a[x].push_back(Edge(y,z));
		a[y].push_back(Edge(x,z));
		id[i]=ID(x,y,a[x].size()-1,a[y].size()-1);
	}
	
	while(q--) {
		int opr;
		read(opr);
		if(opr==1) {
			int x,d;
			read(x),read(d);
			w[x]=d; 
		} else if(opr==2) {
			int e,d;
			read(e),read(d);
			int x=id[e].x,y=id[e].y,p1=id[e].p1,p2=id[e].p2;
			a[x][p1].z=d,a[y][p2].z=d;
		} else {
			dfs1(1,0);
			dfs2(1,0);
			ll ans=0;
			for(int i=1;i<=n;i++) ans^=f[i];
			printf("%lld\n",ans);
		} 
	}
	
	
	return 0;
}

洛谷P3478 [POI2008]STA-Station

给定一个 nn 个点的树，请求出一个结点，使得以这个结点为根时，所有结点的深度之和最大。
一个结点的深度之定义为该节点到根的简单路径上边的数量。

#include<bits/stdc++.h>
using namespace std;

#define read(x) scanf("%d",&x)
#define maxn int(1e6)
#define ll long long

int n;
vector<int> a[maxn+5];

ll sz[maxn+5],d[maxn+5],f[maxn+5];	//子树的大小，对于节点1的深度，答案 

void dfs1(int x,int fa) {
	sz[x]=1;
	for(int i=0;i<a[x].size();i++) {
		int y=a[x][i];
		if(y==fa) continue;
		d[y]=d[x]+1;
		dfs1(y,x);
		sz[x]+=sz[y];
	}
}

void dfs2(int x,int fa) {	//当前x为新根 
	for(int i=0;i<a[x].size();i++) {
		int y=a[x][i];
		if(y==fa) continue;
		f[y]=f[x]+(n-sz[y])-sz[y];
		dfs2(y,x); 
	}
}

int main() {
	read(n);
	for(int i=1;i<n;i++) {
		int x,y;
		read(x),read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	
	d[1]=1;
	dfs1(1,0);
	for(int i=1;i<=n;i++) f[1]+=d[i];
	dfs2(1,0);
	
	int ans=1;
	ll m=-1;
	for(int i=1;i<=n;i++) {
		if(f[i]>m) {
			m=f[i],ans=i;
		}
	}
	
	printf("%d",ans);
	
	return 0;
}