POJ 2378 树状dp

Tree Cutting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3849		Accepted: 2304

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).

Source

USACO 2004 December Silver

 

POJ 2378：

题目意思：

在一棵树上去掉一些点，使得这棵树可以变成一些包含的顶点数目不超过原来的树的一般的一些树，求需要删除的节点，如果有多个节点，按照升序输出;

解题思路：

对树进行宽度优先遍历，如果一个结点的所有以这个节点的子节点为根的树的节点数不超过n/2,并且n减去以这个节点为根节点的树的节点个数小于等于n/2，则这个节点为可以删除的节点；

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=10007;
int dp[maxn];
bool input[maxn];
vector<int> G[maxn];
int n;
void init()
{
    memset(dp,0,sizeof(dp));
    memset(input,0,sizeof(input));
    for(int i=0;i<maxn;i++)
        G[i].clear();
}
void bfs(int point,int father)
{
    dp[point]=1;
    bool flag=true;int num=0;
    for(int i=0;i<G[point].size();i++)
    {
        int v=G[point][i];
        if(v==father) continue;
        bfs(v,point);
        if(dp[v]>n/2) flag=false;
        dp[point]+=dp[v];
    }
   if(flag&&n-dp[point]<=n/2)input[point]=true;
}
int main()
{
   //freopen("in.txt","r",stdin);
    int a,b,root;
    while(~scanf("%d",&n)){
        init();root=-1;
        for(int i=1;i<n;i++)
        {
            scanf("%d %d",&a,&b);
            G[a].push_back(b);
            G[b].push_back(a);
        }
        bfs(1,-1);
        for(int i=1;i<=n;i++)  if(input[i]) printf("%d
",i);
    }
    return 0;
}