Poj1655balancing Act (center of gravity of the tree)

Language:DefaultBalancing ACT Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9418 Accepted: 3941 Description Consider a tree T with N (1 <= n <= 20,000) nodes numbered 1...N. Deleting all node from the tree yields a forest:a Collection of one or more trees. Define the balance of a node to being the size of the largest tree in the forest t created by deleting that node from T. For example, consider the tree: Deleting node 4 yields-trees whose member nodes is {5} and {1,2,3,6,7}. The larger of these and trees have five nodes, thus the balance of node 4 is five. Deleting Node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of the these trees has a nodes, so the balance of node 1 is a. For each input tree, calculate the node, the minimum balance. If multiple nodes has equal balance, output the one with the lowest number. Input The first line of input contains a single integer t (1 <= t <=), the number of test cases. The first line of all test case contains an integer n (1 <= n <= 20,000), the number of congruence. The next N-1 lines each contains and space-separated node numbers that is the endpoints of an edge in the tree. No Edge would be listed twice, and all edges would be listed. Output For each test case, print a line containing-integers, the number of the node with minimum balance and the balance of T Hat node. Sample Input 172 61 21 44 53 73 1 Sample Output 1 2 Source POJ monthly--2004.05.15 IOI 2003 Sample Task

Code

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include < cmath> #include <queue> #include <stack> #include <vector> #include <set> #include <map > #define L (x) (x<<1) #define R (x) (x<<1|1) #define MID (x, y) ((x+y) >>1) #define EPS 1e-8typedef __ Int64 ll;using namespace std; #define N 20005int num[n],dp[n];int head[n],k;int n;struct stud{int to,next;} e[n*2];inline void Add (int u,int v) {e[k].to=v;e[k].next=head[u];head[u]=k++;}    void Dfs (int x,int pre) {dp[x]=0;    Num[x]=1;    int i; for (i=head[x];i!=-1;i=e[i].next) {int to=e[i].to;if (to==pre) Continue;dfs (to,x);DP [X]=max (Dp[x],num[to]); num[x]+= Num[to];} Dp[x]=max (Dp[x],n-num[x]); For example, the tree is 1 2 3 4 5, without this step dp[5] is 0}int main () {int i,j,t;scanf ("%d", &t), while (t--) {K=0;memset (head,-1,sizeof (head)); scanf ("%d", &n), int u,v;for (i=1;i<n;i++) {scanf ("%d%d", &u,&v); add (u,v); add (v,u);} DFS (1,-1); int ans1=1,ans2=dp[1];for (i=2;i<=n;i++) if (Dp[i]<anS2) {ans1=i;ans2=dp[i];} printf ("%d%d\n", ans1,ans2);} return 0;}

Poj1655balancing Act (center of gravity of the tree)