给定一个有$n$个点$m$条边的无向连通图，有$m$次询问，每次询问第$i$条边的权值最大为多少，这张图的所有最小生成树的方案中，一定包含第$i$条边。

先跑一边最小生成树，得到最小生成树，然后依次统计答案，分两种情况：

1、这条边在最小生成树上

假设这条边的边权为$value$，把这条边断开，则分成了两个集合$A,B$，那么有一个性质，从集合$A$连向集合$B$的边，边权一定大于等于$value$，

如果从集合$A$连向集合$B$的边只有一条，则这条边必定存在，答案为$-1$。

如果从集合$A$连向集合$B$的边，边权为$value$的只有一条，则答案为除这条边以外的最小边权 - 1。

否则的话，答案就是$value-1$了 。

2、这条边不在最小生成树上

考虑找到$u->v$路径上权值最大的边$value$，把这条边断开，分成了两个集合$A,B$，同样的有一个性质，从集合$A$连向集合$B$的边，边权一定大于等于$value$，

所以要使$A,B$集合联通，且这条边一定出现，则边权为$value-1$即可。

对第 2 种情况，可以考虑树链剖分，然后线段树维护区间最大值即可，单次询问复杂度$\log n\log n$，考虑如何解决第一种情况。

对第 1 种情况，用线段树维护一个区间最小值跟区间最小值个数，然后单点查询最小值个数即可，边权下放成点权，树链剖分。

#include <bits/stdc++.h>
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1using namespace std;const int N = 2e5 + 10;int head[N], to[N << 1], nex[N << 1], value[N << 1], cnt = 1;int son[N], sz[N], dep[N], fa[N], rk[N], id[N], top[N], tot;int ff[N], vis[N], w[N], ans[N], n, m;struct Res {int u, v, w, id;bool operator < (const Res &t) const {return w < t.w;}
}edge[N];void add(int x, int y, int w) {to[cnt] = y;nex[cnt] = head[x];value[cnt] = w;head[x] = cnt++;
}int find(int rt) {return ff[rt] == rt ? rt : ff[rt] = find(ff[rt]);
}void dfs1(int rt, int f) {fa[rt] = f, dep[rt] = dep[f] + 1, sz[rt] = 1;for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}w[to[i]] = value[i];dfs1(to[i], rt);sz[rt] += sz[to[i]];if (!son[rt] || sz[to[i]] > sz[son[rt]]) {son[rt] = to[i];}}
}void dfs2(int rt, int tp) {rk[++tot] = rt, id[rt] = tot, top[rt] = tp;if (!son[rt]) {return ;}dfs2(son[rt], tp);for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa[rt] || to[i] == son[rt]) {continue;}dfs2(to[i], to[i]);}
}void Kruskal() {sort(edge + 1, edge + m + 1);for (int i = 1; i <= n; i++) {ff[i] = i;}for (int i = 1, sum = 0; i <= m && sum < n; i++) {int u = find(edge[i].u), v = find(edge[i].v);if (u == v) {continue;}ff[u] = v, vis[i] = 1, sum++;add(edge[i].u, edge[i].v, edge[i].w);add(edge[i].v, edge[i].u, edge[i].w);}dfs1(1, 0);dfs2(1, 1);
}int maxn[N << 2], minn[N << 2], lazy[N << 2];void push_up(int rt) {maxn[rt] = max(maxn[ls], maxn[rs]);
}void push_down(int rt) {if (lazy[rt] != 0x3f3f3f3f) {minn[ls] = min(minn[ls], lazy[rt]), minn[rs] = min(minn[rs], lazy[rt]);lazy[ls] = min(lazy[ls], lazy[rt]), lazy[rs] = min(lazy[rs], lazy[rt]);lazy[rt] = 0x3f3f3f3f;}
}void build(int rt, int l, int r) {minn[rt] = lazy[rt] = 0x3f3f3f3f;if (l == r) {maxn[rt] = w[rk[l]];return ;}build(lson);build(rson);push_up(rt);
}void update(int rt, int l, int r, int L, int R, int v) {if (l >= L && r <= R) {lazy[rt] = min(lazy[rt], v);minn[rt] = min(minn[rt], v);return ;}push_down(rt);if (L <= mid) {update(lson, L, R, v);}if (R > mid) {update(rson, L, R, v);}
}int query_max(int rt, int l, int r, int L, int R) {if (l >= L && r <= R) {return maxn[rt];}push_down(rt);int maxn = 0;if (L <= mid) {maxn = max(maxn, query_max(lson, L, R));}if (R > mid) {maxn = max(maxn, query_max(rson, L, R));}return maxn;
}int query_min(int rt, int l, int r, int x) {if (l == r) {return minn[rt];}push_down(rt);if (x <= mid) {return query_min(lson, x);}else {return query_min(rson, x);}
}void update(int u, int v, int w) {while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) {swap(u, v);}update(1, 1, n, id[top[u]], id[u], w);u = fa[top[u]];}if (u != v) {if (dep[u] > dep[v]) {swap(u, v);}update(1, 1, n, id[u] + 1, id[v], w);}
}int query(int u, int v) {int maxn = 0;while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) {swap(u, v);}maxn = max(maxn, query_max(1, 1, n, id[top[u]], id[u]));u = fa[top[u]];}if (u != v) {if (dep[u] > dep[v]) {swap(u, v);}maxn = max(maxn, query_max(1, 1, n, id[u] + 1, id[v]));}return maxn;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &m);for (int i = 1, u, v, w; i <= m; i++) {scanf("%d %d %d", &u, &v, &w);edge[i] = {u, v, w, i};}Kruskal();build(1, 1, n);for (int i = 1; i <= m; i++) {if (!vis[i]) {update(edge[i].u, edge[i].v, edge[i].w);}}for (int i = 1; i <= m; i++) {if (vis[i]) {int u = edge[i].u, v = edge[i].v;if (dep[u] < dep[v]) {swap(u, v);}int cur = query_min(1, 1, n, id[u]);if (cur == edge[i].w) {ans[edge[i].id] = edge[i].w - 1;}else if (cur == 0x3f3f3f3f) {ans[edge[i].id] = -1;}else {ans[edge[i].id] = cur - 1;}}else {ans[edge[i].id] = query(edge[i].u, edge[i].v) - 1;}}for (int i = 1; i <= m; i++) {printf("%d%c", ans[i], i == m ? '\n' : ' ');}return 0;
}