poj2378(dfs) Tree Cutting

Tree Cutting

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3263 Accepted: 1919

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

101 22 33 44 56 77 88 99 103 8

Sample Output

38

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

Memory: 1480 KBTime: 141 MSLanguage: C++Result: Accepted

不多说，这题是前两篇代码的根源，这一个代码用了好几次，没办法这几题实在是太像了

#include<iostream>#include<vector>using namespace std;struct tree{int val,nod;//去掉改点后子节点中包含节点最大的数目和包含节点数vector<int>son;//子节点id}f[10005];int vis[10005],n;void dfs(int x){int i,j;f[x].nod=1;//含本身f[x].val=0;vis[x]=1;for(i=0;i<f[x].son.size();i++){if(!vis[f[x].son[i]]){dfs(f[x].son[i]);f[x].nod+=f[f[x].son[i]].nod;//记录子数及其本身节点数目。f[f[x].son[i]]才是变量if(f[x].val<f[f[x].son[i]].nod)f[x].val=f[f[x].son[i]].nod;}}if(f[x].val<n-f[x].nod)f[x].val=n-f[x].nod;//确定是不是除自己之外最大结点数}int main(){int i,j,k,a,b;while(scanf("%d",&n)!=EOF){memset(vis,0,sizeof(vis));for(i=0;i<n-1;i++){scanf("%d%d",&a,&b);f[b].son.push_back(a);//双向的f[a].son.push_back(b);}dfs(1);//输入的是从1开始的所以从1开始搜索for(i=1;i<=n;i++)//也是从1开始{if(2*f[i].val<=n)//去掉该点后剩下节点里没有包含超过n/2节点的{printf("%d\n",i);}}}return 0;}