Problem:
Problem Statement

You are given a connected undirected graph which has even numbers of nodes.
A connected graph is a graph in which all nodes are connected directly or indirectly by edges.

Your task is to find a spanning tree whose median value of edges' costs is minimum.
A spanning tree of a graph means that a tree which contains all nodes of the graph.

Input

The input consists of multiple datasets.

The format of each dataset is as follows.

n m
s_1 t_1 c_1
...
s_m t_m c_m

The first line contains an even number n (2 \leq n \leq 1,000) and an integer m (n-1 \leq m \leq 10,000).
n is the nubmer of nodes and m is the number of edges in the graph.

Then m lines follow, each of which contains s_i (1 \leq s_i \leq n), t_i (1 \leq s_i \leq n, t_i \neq s_i) and c_i (1 \leq c_i \leq 1,000).
This means there is an edge between the nodes s_i and t_i and its cost is c_i.
There is no more than one edge which connects s_i and t_i.

The input terminates when n=0 and m=0.
Your program must not output anything for this case.

Output

Print the median value in a line for each dataset.

Sample Input

2 1
1 2 5
4 6
1 2 1
1 3 2
1 4 3
2 3 4
2 4 5
3 4 6
8 17
1 4 767
3 1 609
8 3 426
6 5 972
8 1 607
6 4 51
5 1 683
3 6 451
3 4 630
8 7 912
3 7 43
4 7 421
3 5 582
8 4 538
5 7 832
1 6 345
8 2 608
0 0

Output for the Sample Input

5
2
421

Solution in C++ (judge verdict: Accepted)

#include <cstdio>
#include <vector>
#include <cstdlib>
#include <algorithm>

using namespace std;

struct edge {
	int u, v, w;
	edge(int _u = 0, int _v = 0, int _w = 0) : u(_u), v(_v), w(_w) {}
	bool operator<(const edge& o) const {
		return w < o.w;
	}
};

int fa[1005];

int root(int x) {
	return fa[x] = (x == fa[x] ? x : root(fa[x]));
}

int n, m;
vector<edge> e;

int main() {
	while (scanf("%d%d", &n, &m) && n) {
		e.clear();
		for (int i = 0; i <= n; ++i) {
			fa[i] = i;
		}
		int u, v, w;
		for (int i = 0; i < m; ++i) {
			scanf("%d%d%d", &u, &v, &w);
			e.push_back(edge(u, v, w));
		}
		sort(e.begin(), e.end());
		int cnt = 0;
		for (int i = 0; i < m; ++i) {
			int fa1 = root(e[i].u);
			int fa2 = root(e[i].v);
			if (fa1 != fa2) {
				fa[fa1] = fa2;
				if (++cnt == (n + 1)/2) {
					printf("%d\n", e[i].w);
					break;
				}
			}
		}
	}
    return 0;
}