CodeForces 1076E Vasya and a Tree(DFS)

最新推荐文章于 2020-10-18 21:32:53 发布

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探讨在一棵以1号节点为根的树上，如何高效处理一系列操作，即对特定节点及其子树中距离不超过预设值的所有节点进行值更新，最终求得每个节点的总更新值。通过深度优先搜索(DFS)算法与动态规划思想，实现快速求解。

Vasya and a Tree

time limit per test：2 seconds memory limit per test：256 megabytes

Problem Description

Vasya has a tree consisting of n vertices with root in vertex 1. At first all vertices has 0 written on it.

Let $d (i, j)$ be the distance between vertices i and j, i.e. number of edges in the shortest path from i to j. Also, let’s denote k-subtree of vertex x — set of vertices y such that next two conditions are met:

x is the ancestor of y (each vertex is the ancestor of itself);
$d (x, y) \leq k$ .

Vasya needs you to process m queries. The i-th query is a triple v_i, d_i and x_i. For each query Vasya adds value x_i to each vertex from d_i-subtree of v_i.

Report to Vasya all values, written on vertices of the tree after processing all queries.

Input

The first line contains single integer n (1≤n≤3⋅10⁵) — number of vertices in the tree.

Each of next n−1 lines contains two integers x and y (1≤x,y≤n) — edge between vertices x and y. It is guarantied that given graph is a tree.

Next line contains single integer m (1≤m≤3⋅10⁵) — number of queries.

Each of next m lines contains three integers v_i, d_i, x_i (1≤v_i≤n, 0≤d_i≤10⁹, 1≤x_i≤10⁹) — description of the i-th query.

Output

Print n integers. The i-th integers is the value, written in the i-th vertex after processing all queries.

Sample Input

5
1 2
1 3
2 4
2 5
3
1 1 1
2 0 10
4 10 100

Sample Output

1 11 1 100 0

题意

一棵以1号为根，共有n个节点的树，每个节点初始值为0。每次操作可以将以u为根的子树中与其距离不超过k的所有节点都加上x，求所有操作完成后每个节点的值。

题解：

DFS，对于每个节点，假设当前节点u，深度为height。设add[i] 表示深度为i开始需要加的x的和。那个对于从u开始的每操作，DFS向下时，可以令 $a d d [h e i g e h t] + = x$ ， $a d d [h e i g h t + k + 1] - = x$ ，然后递归向下时，设累加变量sum加上add[u]。同时在u递归结束时，需要将add对应的部分进行加减，以消去u的影响。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<ctype.h>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<iterator>
#define dbg(x) cout<<#x<<" = "<<x<<endl;
#define INF 0x3f3f3f3f
#define eps 1e-8
 
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 300100;
const int mod = 1000000007;
LL add[maxn], ans[maxn];
vector<int> g[maxn];
vector<P> que[maxn];
void dfs(int u, int hei, int fa, LL sum);

int main()
{
	int n, m, f, t, i;
	scanf("%d", &n);
	for(i=1;i<n;i++){
		scanf("%d %d", &f, &t);
		g[f].push_back(t);
		g[t].push_back(f);
	}
	scanf("%d", &m);
	for(i=0;i<m;i++)
	{
		int w;
		scanf("%d %d %d", &f, &t, &w);
		que[f].push_back(P(t, w));
	}
	dfs(1, 0, 0, 0);
	for(i=1;i<=n;i++)
	{
		if(i>1)printf(" ");
		printf("%I64d", ans[i]);
	}	
	printf("\n");
	return 0;
}

void dfs(int u, int hei, int fa, LL sum)
{
	int i;
	for(i=0;i<que[u].size();i++)
	{
		int h = min(hei + que[u][i].first+1, maxn-10);
		add[hei] += que[u][i].second;
		add[h] -= que[u][i].second;
	}
	sum += add[hei];
	ans[u] = sum;
	for(i=0;i<g[u].size();i++)
		if(g[u][i] != fa)
			dfs(g[u][i], hei+1, u, sum);
	for(i=0;i<que[u].size();i++)
	{
		int h = min(hei + que[u][i].first+1, maxn-10);
		add[hei] -= que[u][i].second;
		add[h] += que[u][i].second;
	}
}