poj 2378 Tree Cutting

最新推荐文章于 2021-08-10 19:30:55 发布

转载最新推荐文章于 2021-08-10 19:30:55 发布 · 198 阅读

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原文链接：http://www.cnblogs.com/tom987690183/p/3600352.html

本文探讨了一种在特定网络结构中通过移除某些节点来实现网络分裂的方法，目标是确保分裂后的每个子网规模不超过原网络的一半。文章提供了一个具体的算法实现案例，通过深度优先搜索(DFS)算法计算网络分割的有效节点。

Tree Cutting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3525		Accepted: 2083

Description

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

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).

Source

USACO 2004 December Silver

题意：和poj1655几乎是一样的。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<queue>
 6 #include<vector>
 7 #include<algorithm>
 8 using namespace std;
 9 
10 vector<int>Q[10002];
11 int num[10002];
12 int dp[10002][2];
13 bool use[10002];
14 int Queue[10002],len;
15 //dp[i][0] ×Ó½ÚµãµÄ×î´óÊýÁ¿
16 //dp[i][1] Í³¼Æ ×Ó½Úµã¼°×ÔÉí µÄ×Ü¸öÊý
17 
18 void add(int x,int y)
19 {
20     Q[x].push_back(y);
21     num[x]++;
22 }
23 int Max(int x,int y)
24 {
25     return x>y? x:y;
26 }
27 void dfs(int k)
28 {
29     int i,t,cur=0;
30     use[k]=true;
31     dp[k][1]=1;
32     for(i=0;i<num[k];i++)
33     {
34         t=Q[k][i];
35         if(use[t]==true)continue;
36         dfs(t);
37         if(cur<dp[t][1])
38             cur=dp[t][1];
39         dp[k][1]=dp[k][1]+dp[t][1];
40     }
41     dp[k][0]=cur;
42 }
43 int main()
44 {
45     int n;
46     int i,x,y,cur;
47     while(scanf("%d",&n)>0)
48     {
49         for(i=0;i<=n;i++)
50         {
51             Q[i].clear();
52             num[i]=0;
53         }
54         memset(dp,0,sizeof(dp));
55         memset(use,false,sizeof(use));
56 
57         for(i=1;i<n;i++)
58         {
59             scanf("%d%d",&x,&y);
60             add(x,y);
61             add(y,x);
62         }
63         dfs(1);
64         for(i=1,len=0;i<=n;i++)
65         {
66             cur=Max(dp[i][0],n-dp[i][1]);
67             if(cur<=n/2)
68             {
69                 Queue[++len]=i;
70             }
71         }
72         if(len==0)printf("NONE\n");
73         else
74         {
75             sort(Queue+1,Queue+1+len);
76             for(i=1;i<=len;i++)
77                 printf("%d\n",Queue[i]);
78         }
79     }
80     return 0;
81 }

转载于:https://www.cnblogs.com/tom987690183/p/3600352.html