poj2378 树形DP

C - 树形dp

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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).

题目大意：给你一个树，让你删除其中某一个节点，使得剩下的部分节点数不大于n/2，如果满足则输出这个节点，若没有就输出NONE；

思路分析：用链式前向星存图，选择某一个结点作为根节点DFS一遍，将该结点所在的子树的结点数目跑出来，然后在dfs一遍，跑若该节点断开，分开的几段中节点数

最大值DP[i]=max(n-sum[i],max(sum[son]))

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=10000+100;
struct node
{
    int to;
    int next;
};
int sum[maxn];
node edge[2*maxn];
int head[maxn];
int dp[maxn];
bool vis[maxn];//把一个无向 图变成有向图
int tot;
int n;
void add(int x,int y)
{
    edge[tot].to=y;
    edge[tot].next=head[x];
    head[x]=tot++;
}
void init()
{
    memset(head,-1,sizeof(head));
    memset(vis,false,sizeof(vis));
    tot=1;
   int a,b;
   for(int i=1;i<n;i++)//构造图
   {
      scanf("%d%d",&a,&b);
      add(a,b);
      add(b,a);
   }
}
void dfs(int nod)
{
   sum[nod]=1;
   vis[nod]=true;
   for(int i=head[nod];i!=-1;i=edge[i].next)
   {
       int v=edge[i].to;
       if(!vis[v])
       {
           dfs(v);
           sum[nod]+=sum[v];
       }
   }
}
void dfs2(int nod)
{
    dp[nod]=n-sum[nod];
    vis[nod]=true;
    for(int i=head[nod];i!=-1;i=edge[i].next)
    {
      int v=edge[i].to;
      if(!vis[v])
      {
          dfs2(v);
          dp[nod]=max(dp[nod],sum[v]);
      }
    }
}
bool flag=true;
int main()
{
   while(scanf("%d",&n)!=EOF)
   {
       init();
       dfs(1);
       memset(vis,false,sizeof(vis));//重置标记数组
       dfs2(1);
       for(int i=1;i<=n;i++)
       {
           if(dp[i]*2<=n)
           {
               printf("%d
",i);
               flag=false;
           }
       }
       if(flag) printf("NONE
");
   }
}