POJ 1655 Balancing Act （树的重心）

标签：poj

题意：求树的重心裸题。

拓展：树的重心可以在树分治时避免退化链时的最坏时间复杂度，提高时间效率。

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
#define pii (pair<int, int>)
//#pragma comment(linker, "/STACK:1024000000,1024000000") 
using namespace std;  

const int maxn = 30000;
const int INF = 0x3f3f3f3f;
vector<int> G[maxn];
int n, root; //root记录重心 
int des[maxn], bal[maxn], vis[maxn];//des记录以该节点为祖先的后代数，bal记录以该节点为根的最大子树的节点数 

void get_root(int cur) {
	des[cur] = 1;
	bal[cur] = 0;
	vis[cur] = 1;
	int sz = G[cur].size();
	for(int i = 0; i < sz; i++) {
		int u = G[cur][i];
		if(vis[u]) continue;
		get_root(u);
		des[cur] += des[u];
		bal[cur] = max(bal[cur], des[u]);
	}
	bal[cur] = max(bal[cur], n-des[cur]);
	if(bal[cur] < bal[root]) root = cur;
	else if(bal[cur] == bal[root]) root=min(cur, root);
} 

int main() {
	//freopen("input.txt", "r", stdin);
	int T; cin >> T;
	while(T--) {
		scanf("%d", &n);
		for(int i = 1; i <= n; i++) G[i].clear();
		memset(vis, 0, sizeof(vis));
		for(int i = 1; i < n; i++) {
			int u, v; scanf("%d%d", &u, &v);
			G[u].push_back(v);
			G[v].push_back(u);
		}
		root = 0; bal[0] = INF;
		get_root(1);
		cout << root << " " << bal[root] << endl;
	} 
	return 0;
}

POJ 1655 Balancing Act （树的重心）

标签：poj

原文地址：http://blog.csdn.net/u014664226/article/details/47401619