(简单) 树形dp POJ 2378 Tree Cutting

Tree Cutting Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 3398 Accepted: 2004 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 Input10 1 2 2 3 3 4 4 5 6 7 7 8 8 9 9 10 3 8 Sample Output3 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

题意：给出一颗树，如果删除一个节点，那么这棵树会被分为几棵树，问删除哪些节点能使分出来的每颗树的大小都不超过原来的树的大小的一半，并且最大的分出来的树的在所有可行方案中大小最小。
思路：记录每个节点一下的子孙数目，并且记录它的所有儿子形成的子树中最大的子树的大小。 然后。。。就基本做好了
代码：#include<cstdio>#include<algorithm>#include<sstream>#include<set>#include<iostream>#include<map>#include<cmath>#include<string>#include<queue>#include<vector>using namespace std;const int maxn = 10000+10;const int inf = 2000000;#define LL long long
int sum[maxn];int max_tree[maxn];int ptr , n , sz;int ans[maxn];struct Node{ int v; Node *next;}*first[maxn] , edge[2*maxn];
void init(){ ptr = sz = 0; memset(first,0,sizeof(first));}
void add(int x,int y){ edge[++ptr].next = first[x]; edge[ptr].v = y; first[x] = &edge[ptr];}
void dfs(int x,int fa){ Node *p = first[x]; sum[x] = 1; max_tree[0] = 0; while (p) { int v = p->v; if (v!=fa) { dfs(v,x); sum[x] += sum[v]; max_tree[x] = max(max_tree[x],sum[v]); } p = p->next; }}
int main(){ while (scanf("%d",&n)==1) { init(); for (int i = 0 ; i < n-1 ; ++i) { int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x); } dfs(1,-1); int min_tree = inf; for (int i = 1 ; i <= n ; ++i) { int tem = max(max_tree[i],n-sum[i]); if (tem <= (n>>1)) { if (tem<min_tree) { sz = 0; ans[sz++] = i; min_tree = tem; } else if (tem==min_tree) ans[sz++] = i; } } if (sz==0) printf("NONE\n"); else  { for (int i = 0 ; i < sz ; ++i) printf("%d\n",ans[i]); } }}