POJ1655--Balancing Act
来源:互联网 发布:嗑瓜子网络意思 编辑:程序博客网 时间:2024/06/08 12:04
Description
Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T.
For example, consider the tree:
Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.
For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.
For example, consider the tree:
Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.
For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.
Input
The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.
Output
For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.
Sample Input
172 61 21 44 53 73 1
Sample Output
1 2
题意:找一个节点,使得割去该节点后剩下的子树中size最大的最小。
思路:树形DP。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define maxn 40080int dp[maxn];int first[maxn],vv[maxn],nxt[maxn],cnt[maxn];int e,n;void init(){memset(first,-1,sizeof(first));e = 0;}void addedge(int u,int v){vv[e] = v;nxt[e] = first[u];first[u] = e++;}void dfs(int u,int fa){cnt[u] = 1;int tmp = 0;for(int i = first[u];i != -1;i = nxt[i]){int v = vv[i];if(v == fa)continue;dfs(v,u);cnt[u] += cnt[v];tmp = max(tmp,cnt[v]);}dp[u] = max(tmp,n-cnt[u]);}int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&n);init();for(int i = 1;i < n;i++){int u,v;scanf("%d%d",&u,&v);addedge(u,v);addedge(v,u);}dfs(1,0);int ans = 1;for(int i = 2;i <= n;i++){if(dp[ans] > dp[i])ans = i;}printf("%d %d\n",ans,dp[ans]);}return 0;}
0 0
- poj1655(dfs) Balancing Act
- POJ1655--Balancing Act
- 【poj1655】Balancing Act
- POJ1655 Balancing Act
- [POJ1655] Balancing Act
- POJ1655 Balancing Act(树状DP)
- poj1655(Balancing Act + 树形dfs)
- poj1655 Balancing Act 树状dp
- POJ1655:Balancing Act(树形DP)
- POJ1655——Balancing Act
- poj1655 Balancing Act(树形dp)
- 树重心 poj1655 Balancing Act
- poj1655 Balancing Act 【树形DP(很弱)】
- POJ1655 Balancing Act (树的重心)
- POJ1655 Balancing Act 求树的重心
- 【poj1655】Balancing Act 求树的重心
- poj1655 Balancing Act(树形dp)
- 【POJ1655】Balancing Act【树的重心】
- sdut2605 A^X mod P 山东省第四届ACM省赛(打表,快速幂模思想,哈希)
- Java连接MySQL数据库及简单操作代码
- 程序内存空间的分配
- c语言之良好的编程习惯(三)
- stagefright与opencore对比
- POJ1655--Balancing Act
- Linux Shell 脚本应用——数值运算及处理
- 【Linux视频教程-韩顺平】入门
- 结构体内存对齐
- iOS编程基础: Hello World App是如何运作的?
- RBAC(Role-Based Access Control,基于角色的访问控制)
- 二分匹配
- sub技巧
- 字符串编程问题:求连续出现最长字符串问题(js实现)