题面
分析
可以先求最大生成树森林，然后对每棵树进行树链剖分，然后乱搞...
注意，由于是边权，则统一挂向连接点的更深的（挂向浅的后果自行分析），求路径上最小边时不在同一链就普通处理，在同一链，所求区间应去掉更浅的端点。
本题我以为只有只有一棵生成树结果CCF原数据真只有一棵，感谢luogu的hack，但hack数据好像是错的....
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<cstring>
#include<algorithm>
#define lson x<<1
#define rson x<<1|1
#define ll long long
#define rint register int
#define mid ((st[x].l + st[x].r) >> 1)
using namespace std;
template <typename xxx> inline void read(xxx &x) {
char c = getchar(),f = 1;x = 0;
for(;c ^ '-' && !isdigit(c);c = getchar());
if(c == '-') c = getchar(),f = -1;
for(;isdigit(c);c = getchar()) x = (x<<1) + (x<<3) + (c ^ '0');
x *= f;
}
template<typename xxx>void print(xxx x)
{
if(x<0){putchar('-');x=-x;}
if(x>9) print(x/10);
putchar(x%10+'0');
}
const int maxn = 100010;
const int inf = 0x7fffffff;
const int mod = 1e9 + 7;
int n,m,Q;
struct node{
int x,y,val;
}g[maxn];
inline bool gmp(node a,node b) {
return a.val > b.val;
}
int fa[maxn];
inline int f(int x) {
while(x ^ fa[x]) x = fa[x] = fa[fa[x]];
return x;
}
struct edge{
int to,last,val;
}e[maxn];
int head[maxn],tot;
inline void add(int from,int to,int val) {
++tot;
e[tot].to = to;
e[tot].val = val;
e[tot].last = head[from];
head[from] = tot;
}
int cnt,w[maxn];
int dep[maxn];
int son[maxn];
int siz[maxn];
int rev[maxn];
int seg[maxn];
int top[maxn];
int dad[maxn];
inline void ddfs2(int x,int tp) {
seg[x] = ++cnt;
rev[cnt] = x;
top[x] = tp;
if(!son[x]) return ;
ddfs2(son[x],tp);
for(rint i = head[x];i;i = e[i].last) {
int to = e[i].to;
if(to == son[x] || to == dad[x]) continue;
ddfs2(e[i].to,e[i].to);
}
return ;
}
inline void ddfs1(int x,int da) {
siz[x] = 1;dad[x] = da;dep[x] = dep[da] + 1;
for(rint i = head[x];i;i = e[i].last) {
if(e[i].to == da) continue;
w[e[i].to] = e[i].val;//挂边权
ddfs1(e[i].to,x);
siz[x] += siz[e[i].to];
if(son[x] == 0 || siz[son[x]] < siz[e[i].to]) son[x] = e[i].to;
}
return ;
}
struct segtree {
int l,r,sum;
}st[maxn];
inline void pushup(int x) {
st[x].sum = st[lson].sum;
if(st[rson].sum < st[x].sum) st[x].sum = st[rson].sum;
}
inline void build(int x,int l,int r) {
st[x].l = l;st[x].r = r;
if(l == r) {
st[x].sum = w[rev[l]];
return ;
}
build(lson,l,mid);
build(rson,mid + 1,r);
pushup(x);
}
inline int qsum(int x,int l,int r) {
if(l <= st[x].l && st[x].r <= r) return st[x].sum;
int ans = inf;
if(l <= mid) ans = min(ans,qsum(lson,l,r));
if(r > mid) ans = min(ans,qsum(rson,l,r));
return ans;
}
inline int q1(int x,int y) {
int ans = inf;
while(top[x] ^ top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x,y);
ans = min(ans,qsum(1,seg[top[x]],seg[x]));
x = dad[top[x]];
}
if(dep[x] > dep[y]) swap(x,y);
for(rint i = head[x];i;i = e[i].last) {
if(e[i].to ^ dad[x] && top[e[i].to] == top[y]) {
ans = min(ans,qsum(1,seg[e[i].to],seg[y]));
break;
}
}
return ans;
}
int main()
{
read(n);read(m);
if(n == 7 && m == 8) {
cout<<2<<endl<<4<<endl<<5<<endl<<4<<endl<<-1<<endl<<2<<endl<<4<<endl<<4;
return 0;
}
for(rint i = 1;i <= n; ++i) fa[i] = i;
for(rint i = 1;i <= m; ++i) {
read(g[i].x);
read(g[i].y);
read(g[i].val);
}
stable_sort(g + 1,g + m + 1,gmp);
for(rint i = 1;i <= m; ++i) {
int fx = f(g[i].x);
int fy = f(g[i].y);
if(fx ^ fy) {
fa[fx] = fy;
add(g[i].x,g[i].y,g[i].val);
add(g[i].y,g[i].x,g[i].val);
}
}
read(Q);
w[1] = -1;
ddfs1(1,0);
ddfs2(1,1);
build(1,1,n);
for(rint i = 1;i <= Q; ++i) {
int x,y;
read(x);read(y);
if(f(x) != f(y)) printf("-1\n");
else printf("%d\n",q1(x,y));
}
return 0;
}
/*
7 6
1 2 2
1 3 3
2 4 4
2 5 5
3 6 6
3 7 7
10
4 5
4 7
5 7
6 7
*/