zoukankan      html  css  js  c++  java
  • POJ 1655 Balancing Act(求树的重心)

    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. 

    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.
     
    题目大意:给一棵树,问删掉哪个结点后,剩下的树的最大结点数最小。
    思路:DFS一次即可求出所有点的子树大小size,顺便算出每个点最大的子树maxSize。那么对于每个点删掉之后,剩下的树的最大结点数就是max(maxSize, n - size),前面就是它的所有子树的最大size,后面就是删掉这个点后,它父亲所在的树的大小。
    在研究树的分治之前先来补一条水题。。。这也算DP- -?
     
    代码(47MS):
     1 #include <cstdio>
     2 #include <iostream>
     3 #include <algorithm>
     4 #include <cstring>
     5 using namespace std;
     6 
     7 const int MAXN = 20010;
     8 const int MAXE = 40010;
     9 const int INF = 0x7fff7fff;
    10 
    11 int head[MAXN], size[MAXN], maxSize[MAXN], f[MAXN];
    12 int to[MAXE], next[MAXE];
    13 int n, ecnt;
    14 
    15 void init() {
    16     memset(head, -1, sizeof(head));
    17     ecnt = 0;
    18 }
    19 
    20 void add_edge(int u, int v) {
    21     to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
    22     to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
    23 }
    24 
    25 void dfs(int u) {
    26     maxSize[u] = 0;
    27     size[u] = 1;
    28     for(int p = head[u]; ~p; p = next[p]) {
    29         int &v = to[p];
    30         if(v == f[u]) continue;
    31         f[v] = u;
    32         dfs(v);
    33         size[u] += size[v];
    34         maxSize[u] = max(maxSize[u], size[v]);
    35     }
    36 }
    37 
    38 int main() {
    39     int T;
    40     scanf("%d", &T);
    41     while(T--) {
    42         init();
    43         scanf("%d", &n);
    44         int u, v;
    45         for(int i = 1; i < n; ++i) {
    46             scanf("%d%d", &u, &v);
    47             add_edge(u, v);
    48         }
    49         dfs(1);
    50         int pos, maxd = INF;
    51         for(int i = 1; i <= n; ++i) {
    52             if(max(maxSize[i], n - size[i]) < maxd) {
    53                 pos = i;
    54                 maxd = max(maxSize[i], n - size[i]);
    55             }
    56         }
    57         printf("%d %d
    ", pos, maxd);
    58     }
    59 }
    View Code
  • 相关阅读:
    Java在处理大数据的时候一些小技巧
    大并发处理解决方案
    数据库SQL优化大总结之 百万级数据库优化方案
    DotNet中的计时器线程计时器
    System.Threading.Timer的使用技巧
    Asp.net Mvc 请求是如何到达 MvcHandler的——UrlRoutingModule、MvcRouteHandler分析,并造个轮子
    C#-结构
    @Html.ActionLink(),@Html.Raw(),@Url.Action()等
    bootstarpt-table小结
    input[ type="file"]上传文件问题
  • 原文地址:https://www.cnblogs.com/oyking/p/3408852.html
Copyright © 2011-2022 走看看