Poj 2378 Tree Cutting (DP _ Tree DP)

Given a tree with n nodes, if you delete a node so that the number of the remaining largest branch nodes is smaller than half of the total number of nodes, the deletion will burst, total number of deleted methods.

Solution: Tree DP. the first deep search record the total number of child nodes (including the current node) from the current node (including the current node), the first time the pre-processing is calculated, the complexity is O (N ), the second time, we use the first result to find the maximum number of nodes in each branch. There are two kinds of branches: one is the Child Branch, this is the branch from the current node to the Father's Day point (total number of N-dp [cur]), so that you can calculate it again N times.

Test data:
10
1 2
2 3
3 4
4 5
6 7
7 8
8 9
9 10
3 8

Code:
[Html]
# Include <stdio. h>
# Include <string. h>
# Include <algorithm>
Using namespace std;
# Deprecision MAX 110000
# Define max (a, B) (a)> (B )? (A) :( B)
 
 
Struct node {
 
Int v;
Node * next;
} * Head [MAX], tree [MAX];
Int n, m, ptr, dp [MAX], ans [MAX], cnt;
 
 
Void Initial (){
 
Cnt = ptr = 0;
Memset (dp, 0, sizeof (dp ));
Memset (head, NULL, sizeof (head ));
}
Void AddEdge (int x, int y ){
 
Tree [ptr]. v = y;
Tree [ptr]. next = head [x], head [x] = & tree [ptr ++];
Tree [ptr]. v = x;
Tree [ptr]. next = head [y], head [y] = & tree [ptr ++];
}
Void Dfs_Ini (int s, int pa ){
 
Dp [s] = 1;
Node * p = head [s];
While (p! = NULL ){
 
If (p-> v! = Pa ){
 
Dfs_Ini (p-> v, s );
Dp [s] + = dp [p-> v];
}
P = p-> next;
}
}
Void Dfs_Solve (int son, int pa ){
 
Int I, j, tp, tot = 0;
Node * p = head [son];

 
While (p! = NULL ){
 
If (p-> v! = Pa ){
 
Dfs_Solve (p-> v, son );
Tp = dp [p-> v];
Tot = max (tot, tp );
}
P = p-> next;
}
If (n-dp [son] <= n/2 & tot <= n/2)
Cnt ++, ans [cnt] = son;
}
 
 
Int main ()
{
Int I, j, k, a, B;
 
 
While (scanf ("% d", & n )! = EOF ){
 
Initial ();
For (I = 1; I <n; ++ I ){
 
Scanf ("% d", & a, & B );
AddEdge (a, B );
}
 
 
Dfs_Ini (); // the first in-depth search to record the total number of children of the current node
Dfs_Solve (); // update the answer
Sort (ans + 1, ans + 1 + cnt); // output in Lexicographic Order
For (I = 1; I <= cnt; ++ I)
Printf ("% d \ n", ans [I]);
If (cnt = 0) printf ("NONE \ n ");
}
}

Author: woshi250hua