Poj 1655 balancing act tree center of gravity

Give a tree. After removing a vertex, the tree will become a Unicom block. Find the smallest number of blocks formed after which point is removed.

Idea: tree-like DP. Find the size of the Child tree with each node as the root through a deep search. After removing the node, the Child tree will become the child trees of the node, with the remaining parts, find the maximum number of these blocks, that is, remove the ANS at this point, and then update the total ans.

This question is actually the focus of the tree.

Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 20010using namespace std;int cases;int points;int head[MAX],total;int next[MAX << 1],aim[MAX << 1];int p_ans,ans;inline void Initialize();inline void Add(int x,int y);int DFS(int x,int last);int main(){for(cin >> cases;cases; --cases) {scanf("%d",&points);Initialize();for(int x,y,i = 1;i < points; ++i) {scanf("%d%d",&x,&y);Add(x,y),Add(y,x);}DFS(1,0);printf("%d %d\n",p_ans,ans);}return 0;}inline void Initialize(){ans = 0x7f7f7f7f,total = 0;memset(head,0,sizeof(head));}inline void Add(int x,int y){next[++total] = head[x];aim[total] = y;head[x] = total;}int DFS(int x,int last){int max_size = 0,size = 1;for(int i = head[x];i;i = next[i]) {if(aim[i] == last)continue;int temp = DFS(aim[i],x);max_size = max(max_size,temp);size += temp;}max_size = max(max_size,points - size);if(max_size < ans)ans = max_size,p_ans = x;return size;}

Poj 1655 balancing act tree center of gravity