车

思路：对于每一次询问x y，我们要求的是x到y之间的一条路径，这条路径满足路径的权值最小值在所有路径中最大，所以我们可以直接构造最大生成树，在这颗树上求解，之后直接用lca求解即可。

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>
#include <cstring>
#include <bitset>
#include <complex>
#include <list>
#include <unordered_map>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int mod = 1000000007;

struct node { int x, y, z,flag; }ea[51111];
int fa[11111], n, m, ans,q;
int find(int x) {
	if (x == fa[x])return x;
	return fa[x] = find(fa[x]);
}

vector<pair<int,int>> e[11111];
int d[11111], f[11111][20],t,mf[11111][20];

void bfs(int s) {
	queue<int> q; q.push(s); d[s] = 1;
	while (q.size()) {
		int x = q.front(); q.pop();
		for (int i = 0; i < e[x].size(); i++) {
			int y = e[x][i].first, z = e[x][i].second;
			if (d[y])continue;
			d[y] = d[x] + 1;
			f[y][0] = x; mf[y][0] = z;
			for (int j = 1; j <= t; j++)
				f[y][j] = f[f[y][j - 1]][j - 1];
			for (int j = 1; j <= t; j++)
				mf[y][j] = min(mf[y][j - 1], mf[f[y][j - 1]][j - 1]);
			q.push(y);
		}
	}
}

int lca(int x, int y) {
	if (d[x] > d[y])swap(x, y);
	int ans = INF;
	for (int i = t; i >= 0; i--)
		if (d[f[y][i]] >= d[x]) 
			ans = min(ans, mf[y][i]), y = f[y][i];
	if (x == y)return ans;
	for (int i = t; i >= 0; i--)
		if (f[x][i] != f[y][i])
			ans = min(ans, mf[x][i]), ans = min(ans, mf[y][i]),
			x = f[x][i], y = f[y][i];
	
	return min(ans, min(mf[x][0],mf[y][0]));
}

signed main() {
	memset(mf, INF, sizeof(mf));
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i++)
		scanf("%d%d%d", &ea[i].x, &ea[i].y, &ea[i].z);
	sort(ea + 1, ea + 1 + m, [](const node& a, const node&b) { return a.z > b.z; });
	for (int i = 1; i <= n; i++)fa[i] = i;
	for (int i = 1; i <= m; i++) {
		int x = find(ea[i].x), y = find(ea[i].y);
		if (x == y)continue;
		fa[x] = y, ea[i].flag = 1;
	}
	for (int i = 1; i <= m; i++) {
		if (ea[i].flag) {
			int x = ea[i].x, y = ea[i].y, z = ea[i].z;
			e[x].push_back(make_pair(y, z));
			e[y].push_back(make_pair(x, z));
		}
	}
	t = (log(n) / log(2)) + 1;
	for (int i = 1; i <= n; i++) 
		if (d[i] == 0)bfs(i);

	scanf("%d", &q);
	for (int i = 1,x,y; i <= q; i++) {
		scanf("%d%d", &x, &y);
		if (find(x) != find(y))printf("-1\n");
		else printf("%d\n", lca(x, y));
	}

	return 0;
}