AIM Tech Round 3 (Div. 1) C. Centroids(树形DP)
来源:互联网 发布:js购物车数量合计 编辑:程序博客网 时间:2024/05/16 05:09
Tree is a connected acyclic graph. Suppose you are given a tree consisting ofn vertices. The vertex of this tree is calledcentroid if the size of each connected component that appears if this vertex is removed from the tree doesn't exceed.
You are given a tree of size n and can perform no more than one edge replacement.Edge replacement is the operation of removing one edge from the tree (without deleting incident vertices) and inserting one new edge (without adding new vertices) in such a way that the graph remains a tree. For each vertex you have to determine if it's possible to make it centroid by performing no more than one edge replacement.
The first line of the input contains an integer n (2 ≤ n ≤ 400 000) — the number of vertices in the tree. Each of the nextn - 1 lines contains a pair of vertex indicesui andvi (1 ≤ ui, vi ≤ n) — endpoints of the corresponding edge.
Print n integers. The i-th of them should be equal to 1 if thei-th vertex can be made centroid by replacing no more than one edge, and should be equal to0 otherwise.
31 22 3
1 1 1
51 21 31 41 5
1 0 0 0 0
In the first sample each vertex can be made a centroid. For example, in order to turn vertex1 to centroid one have to replace the edge(2, 3) with the edge (1, 3).
题意:给你一棵树,对于每个顶点,问最多更改一条边后能否把它变为树的重心。
分析:树形DP,计算向下和向上最多能找到的点数不大于n/2的子树,注意如果一个点不是重心,那么它最多只有一个儿子的节点数超过n/2,我们计算完所有辅助值后用改超标儿子的度数减去它的最大不超标子树,判断此时的值是否还大于n/2即可。对于这种树形DP记录最优解和次优解辅助转移是一种常见技巧。
#include<iostream>#include<string>#include<algorithm>#include<cstdlib>#include<cstdio>#include<set>#include<map>#include<vector>#include<cstring>#include<stack>#include<cmath>#include<queue>#include <unordered_map>#define INF 0x3f3f3f3f#define eps 1e-9 #define MOD 1000000007 #define MAXN 400005using namespace std;typedef long long ll;vector<int> G[MAXN];int n,sz[MAXN],up[MAXN],Down[MAXN],max1[MAXN],max2[MAXN],ans[MAXN];void update(int val,int &x,int &y){if(y < val) y = val;if(x < y) swap(x,y);}void dfs1(int u,int fa){sz[u] = 1;for(int v : G[u]) if(v != fa) { dfs1(v,u); sz[u] += sz[v]; update(Down[v],max1[u],max2[u]); } Down[u] = max1[u];if(sz[u] <= n/2) Down[u] = sz[u];}void dfs2(int u,int fa){if(fa >= 1){up[u] = max(up[u],up[fa]);if(Down[u] != max1[fa]) up[u] = max(up[u],max1[fa]);else up[u] = max(up[u],max2[fa]);if(n - sz[u] <= n/2) up[u] = n - sz[u];}int tmp = 0;for(int v : G[u])if(v != fa) {dfs2(v,u);if(sz[v] > n/2) tmp = v;}if(n - sz[u] > n/2) tmp = -1;if(!tmp) return;if(tmp > 0){if(sz[tmp] - Down[tmp] > n/2) ans[u] = 1;return;} if(n - sz[u] - up[u] > n/2) ans[u] = 1;}int main(){scanf("%d",&n);for(int i = 1;i < n;i++){int x,y;scanf("%d%d",&x,&y);G[x].push_back(y);G[y].push_back(x);}dfs1(1,-1);dfs2(1,-1);for(int i = 1;i <= n;i++) printf("%d ",ans[i]^1);}
- AIM Tech Round 3 (Div. 1) C. Centroids(树形DP)
- AIM Tech Round 3 (Div. 1) C. Centroids(树形dp)
- AIM Tech Round 3 (Div. 1)-C. Centroids
- AIM Tech Round 3 (Div. 2) E. Centroids (树形dp)
- AIM Tech Round 3 (Div. 2) E. Centroids (树形dp) ★ ★ ★
- AIM Tech Round 3 (Div. 1) C. Centroids(每个点能否删掉一条边再添加一条边使得这个点成为重心)
- AIM Tech Round Div 1
- AIM Tech Round (Div. 1) B. Array GCD(数论+dp)
- 【DP】AIM Tech Round (Div. 2) D
- AIM Tech Round 3 (Div. 2) C.Letters Cyclic Shift
- AIM Tech Round 3 (Div. 2) C(贪心)
- CF708A(AIM Tech Round 3 (Div. 2) - C)
- AIM Tech Round 4 (Div. 2) C
- AIM Tech Round 3 (Div. 1)-B. Recover the String
- AIM Tech Round 3 (Div. 2)
- AIM Tech Round 3 (Div. 2)
- AIM Tech Round 3 (Div. 2)
- 【codeforces】AIM Tech Round 3 (Div. 2)
- NetBeans 的安装与简单实用(及注意事项)
- 算法竞赛入门经典第二版第一章语言篇
- reverse vowels of a string (leetcode 345) java
- 软考网络工程师--局域网和城域网
- JS判断手机端是否安装了某个客户端APP应用
- AIM Tech Round 3 (Div. 1) C. Centroids(树形DP)
- c++ stl sort
- js怎么实现⻚⾯挑转到指定 URL
- R语言设置图片标题,坐标轴的颜色
- Linux基础学习--查询数据man 后面的数字意义(1)
- C/C++ socket编程教程之一:socket是什么
- ubuntu 全局环境变量和局部环境变量的设置
- Python 学习笔记之八——输入和输出
- java 选择排序