zoukankan      html  css  js  c++  java
  • poj 2378 Tree Cutting 树形dp

    After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible cost, he sued Bessie to mitigate his losses. 

    Bessie, feeling vindictive, decided to sabotage Farmer John's network by cutting power to one of the barns (thereby disrupting all the connections involving that barn). When Bessie does this, it breaks the network into smaller pieces, each of which retains full connectivity within itself. In order to be as disruptive as possible, Bessie wants to make sure that each of these pieces connects together no more than half the barns on FJ. 

    Please help Bessie determine all of the barns that would be suitable to disconnect.

    Input

    * Line 1: A single integer, N. The barns are numbered 1..N. 

    * Lines 2..N: Each line contains two integers X and Y and represents a connection between barns X and Y.

    Output

    * Lines 1..?: Each line contains a single integer, the number (from 1..N) of a barn whose removal splits the network into pieces each having at most half the original number of barns. Output the barns in increasing numerical order. If there are no suitable barns, the output should be a single line containing the word "NONE".

    Sample Input

    10
    1 2
    2 3
    3 4
    4 5
    6 7
    7 8
    8 9
    9 10
    3 8

    Sample Output

    3
    8

    Hint

    INPUT DETAILS: 

    The set of connections in the input describes a "tree": it connects all the barns together and contains no cycles. 

    OUTPUT DETAILS: 

    If barn 3 or barn 8 is removed, then the remaining network will have one piece consisting of 5 barns and two pieces containing 2 barns. If any other barn is removed then at least one of the remaining pieces has size at least 6 (which is more than half of the original number of barns, 5).
     
     1 #include <cstdio>
     2 #include <algorithm>
     3 #include <cstring>
     4 using namespace std;
     5 const int maxn = 1e5 + 10;
     6 const int INF = 0x7fffffff;
     7 int n, dp[maxn][2], head[maxn], tot, ans[maxn];
     8 struct node {
     9     int v, next;
    10 } edge[maxn];
    11 void init() {
    12     tot = 0;
    13     memset(head, -1, sizeof(head));
    14 }
    15 void add(int u, int v) {
    16     edge[tot].v = v;
    17     edge[tot].next = head[u];
    18     head[u] = tot++;
    19     edge[tot].v = u;
    20     edge[tot].next = head[v];
    21     head[v] = tot++;
    22 }
    23 int k = 0;
    24 int solve(int x, int fa) {
    25     int sum = 1, maxs = -1;
    26     for (int i = head[x] ; i != -1 ; i = edge[i].next) {
    27         int v = edge[i].v;
    28         if (v == fa) continue;
    29         int cost = solve(v, x);
    30         if (cost > maxs) maxs = cost;
    31         sum += cost;
    32     }
    33     if (n - sum > maxs) maxs = n - sum;
    34     if (maxs <= n / 2) ans[k++] = x;
    35     return sum;
    36 }
    37 int main() {
    38     while(scanf("%d", &n) != EOF) {
    39         init();
    40         for (int i = 0 ; i < n - 1 ; i++) {
    41             int u, v;
    42             scanf("%d%d", &u, &v);
    43             add(u, v);
    44         }
    45         solve(1, -1);
    46         if (k == 0) printf("NONE
    ");
    47         else {
    48             sort(ans, ans + k);
    49             for (int i = 0 ; i < k ; i++)
    50                 printf("%d
    ", ans[i]);
    51         }
    52     }
    53     return 0;
    54 }
  • 相关阅读:
    JavaScript HTML DOM
    Java数组声明、初始化
    如何破解MyEclipse 10.x
    SpringBoot框架中解决日期展示问题
    spring boot集成mybatis-plus插件进行自定义sql方法开发时报nested exception is org.apache.ibatis.binding.BindingException: Invalid bound statement (not found):
    Spring boot启动时报 java.sql.SQLException: java.lang.ClassCastException: java.math.BigInteger cannot be cast to java.lang.Long错误
    springboot 配置mybatis打印sql
    解决使用Navicat等工具进行连接登录mysql的1130错误,无法使用Ip远程连接的问题(mysql为8.0版本)
    解决使用Navicat等工具进行连接登录mysql的1521错误,(mysql为8.0版本)
    【转载】VUE的背景图引入
  • 原文地址:https://www.cnblogs.com/qldabiaoge/p/9345710.html
Copyright © 2011-2022 走看看