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Tree Cutting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 892 Accepted: 502

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

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).

题解:
1.构图的时候我们可以用邻接表表示防止MLE(常识), 我很懒直接不要时间的用了vector建议自己写邻接表。
2.题目所给一定是个树n个点n-1条边,题意是在原树上找到所有与其相连的所有子树不大于n/2的节点,按字典序输出!
3.一次O(n)DFS解决所有问题,边为无向边,从任一点开始DFS都可以!按照DFS的原理开始搜索并同时记录子树的值。
4.核心代码:
int DFS(int x){
    
int flag=0,tot=0;
    
for(int i=0;i<v[x].size();i++{
        
if(!mark[v[x][i]]){
            mark[v[x][i]]
=true;
            
int t=DFS(v[x][i]);
            tot
+=t;
            
if(t>n/2)flag=1;
        }

    }

    
if(n-tot-1>n/2)flag=1;
    
if(!flag)now[sign++]=x;
    
return tot+1;
}
posted on 2009-04-12 17:15 KNIGHT 阅读(118) 评论(0)  编辑 收藏 引用

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