题意：给出一张n个点m条边的无向图， 每条边有一个危险度，有q个询问， 每次给出两个点s、t，找一条路， 使得路径上的最大危险度最小。
做法：首先考虑到需要最大危险度最小，那么对于一个无向图很容易想到最小生成树，其次如果我生成最小生成树之后进行LCA更新其路径最大值，那么即可解决本题。
/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
//#pragma GCC optimize("Ofast")
#pragma comment(linker, "/STACK:102400000,102400000")
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx)
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;
#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=5e4+5;
//const int maxx=1e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}
inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}
void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;
int up[maxn][21];
int dep[maxn],dis[maxn];
int cnt, head[maxn];
int maxx[maxn][21];
int n, m, q;
struct node {
int to, next, w;
}e[maxn<<1];
void init() {
me(head,-1);
me(dis,0);
me(up,0);
me(dep,0);
me(maxx, 0);
cnt = 0;
}
void add(int u, int v, int w) {
e[cnt] = node{v, head[u], w};
head[u] = cnt++;
}
void dfs(int u,int fa,int d) {
dep[u] = d + 1;
for(int i = 1 ; i < 20 ; i ++) {
up[u][i] = up[up[u][i-1]][i-1];
maxx[u][i] = max(maxx[up[u][i-1]][i-1], maxx[u][i-1]);
}
for(int i = head[u] ; ~i ; i = e[i].next) {
int to = e[i].to;
if(to == fa) continue;
dis[to] = dis[u] + e[i].w;
up[to][0] = u;
maxx[to][0] = e[i].w;
dfs(to, u, d+1);
}
}
int LCA_BZ(int u,int v) {
int mx = 0;
if(dep[u] < dep[v]) swap(u,v);
int k = dep[u] - dep[v];
for(int i = 19 ; i >= 0 ; i --) {
if((1<<i) & k) {
mx = max(mx, maxx[u][i]);
u = up[u][i];
}
}
if(u != v) {
for(int i = 19 ; i >= 0 ; i --) {
if(up[u][i] != up[v][i]){
mx = max(mx, maxx[u][i]);
mx = max(mx, maxx[v][i]);
u = up[u][i];
v = up[v][i];
}
}
mx=max(mx,maxx[u][0],maxx[v][0]);
u=up[u][0];
}
return mx;
//return up[u][0];
}
//int kth(int u, int v, int k) { // 算u到v的路径的第k个结点.
// --k; int f = LCA_BZ(u, v);
// if (dep[u] - dep[f] < k) {
// k = dep[v] - dep[f] - k + dep[u] - dep[f];
// swap(u, v);
// }
// for (int i = 19 ; i >= 0 ; i --) {
// if (k & (1 << i)) u = up[u][i];
// }
// return u;
//}
struct edge {
int x,y;
LL cost;
}Q[maxn<<1];
int fa[maxn];
void init1() {
for(int i=0;i<=n;i++) fa[i]=i;
me(Q,0);
}
int fi(int x) {
return fa[x]==x?x:fa[x]=fi(fa[x]);
}
void unio(int x,int y) {
int p1=fi(x),p2=fi(y);
if(p1!=p2) fa[p1]=p2;
}
bool cmp(edge a,edge b) {
return a.cost<b.cost;
}
void Kru() {
sort(Q+1,Q+m+1,cmp);
int sum=0;
for(int i=1;i<=m;i++) {
if(fi(Q[i].x)==fi(Q[i].y)) continue;
sum++;
unio(Q[i].x,Q[i].y);
add(Q[i].x,Q[i].y,Q[i].cost);
add(Q[i].y,Q[i].x,Q[i].cost);
if(sum>=n-1) break;
}
}
//
//void solve() {
// scanf("%d%d",&n,&m);
// init();
// for(int i = 1 ; i < n ; i ++) {
// int u, v, w;
// scanf("%d%d%d", &u, &v, &w);
// add(u, v, w); add(v, u, w);
// }
// //maxx[1][0] = 0;
// dfs(1, -1, 0);
// while(m--) {
// int u, v;
// scanf("%d%d",&u, &v);
// int res = dis[u] + dis[v] - 2 * dis[LCA_BZ(u,v)];
// printf("%d\n", res);
// }
//}
void solve() {
int f=0;
W(s_2(n,m)!=EOF) {
if(f++) puts("");
init();
init1();
FOR(1,m,i) {
int u,v,w;
s_3(u,v,w);
Q[i]=(edge){u,v,w};
}
Kru();
up[1][0]=1;
maxx[1][0]=INF;
dfs(1,-1,0);
s_1(q);
W(q--) {
int u,v;
s_2(u,v);
int res=LCA_BZ(u,v);
print(res);
}
}
}
int main() {
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++) {
//printf("Case #%d: ",cas);
solve();
}
}