（POJ 2378 Tree Cutting）树型DP + 删点

Tree Cutting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4512 Accepted: 2754
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

/*2017/3/11time： 485MS题意：有一棵有n个节点的树，问你删除哪些节点后，使剩下的子树中节点数最大的值最小？分析：（这是之前POJ 1741 tree这一题的一个子问题）题目不难，先我们求出去掉每个节点后的最大值，然后找出其中的最小值，最后遍历一遍就可在找出删除每个节点后的最大值，我们用dfs_node()首先求出以每个节点为根的子树的节点数目nodes[i]然后删除i节点后的最大值就等于：以他每个孩子节点为根的子树的节点数目的最大值，和n - nodes[i] 中的大者关于上面的问题我们直接dfs一遍更新即可*/#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>using namespace std;const int maxn = 50010;struct edge{    int v,next;}edges[2*maxn];int head[maxn],e,n;int nodes[maxn],maxnum[maxn];void addedges(int u,int v){    edges[e].v = v;    edges[e].next = head[u];    head[u] = e++;}void dfs_subnode(int u,int f){    nodes[u] = 1;    for(int i=head[u];i!=-1;i=edges[i].next)    {        int v = edges[i].v;        if(v == f) continue;        dfs_subnode(v,u);        nodes[u] += nodes[v];    }}void dfs_up(int u,int f){    maxnum[u] = n - nodes[u];    for(int i=head[u];i!=-1;i=edges[i].next)    {        int v = edges[i].v;        if(v == f) continue;        if(maxnum[u] < nodes[v]) maxnum[u] = nodes[v];        dfs_up(v,u);    }}int main(){    int u,v;    while(scanf("%d",&n)!=EOF)    {        memset(head,-1,sizeof(head));        e = 0;        for(int i=0;i<n-1;i++)        {            scanf("%d%d",&u,&v);            addedges(u,v);            addedges(v,u);        }        dfs_subnode(1,0);        dfs_up(1,0);        int minans = maxn;        for(int i=1;i<=n;i++) if(minans > maxnum[i])            minans = maxnum[i];        int t = 0;        for(int i=1;i<=n;i++) if(maxnum[i] == minans)        {            t++;            if(t == 1) printf("%d",i);            else printf(" %d",i);        }        printf("\n");    }    return 0;}