[LuoGu2245]星际导航

Description

[LuoGu2245]星际导航

Solution

好像叫作最小瓶颈路问题
就是把原图求最小生成树，然后在树上跑( ext{LCA})的最大值
如果不在一个一颗树上，就输出( ext{impossible})
玄学证明不会

Code

#include<bits/stdc++.h>
using namespace std;
#define N 800010
struct rec {
	int nxt, ver, val;
} t[N];
struct Rec {
	int u, v, w;
} e[N];
int n, m, q, cnt, tot, u, v;
int fa[N], head[N], deep[N], f[N][21], h[N][21];
inline int read() {
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}
int find(int x) {
	if (fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}
inline void Merge(int x, int y) {
	fa[find(x)] = find(y);
}
inline bool cmp(Rec x, Rec y) {
	return x.w < y.w;
}
inline void add(int u, int v, int w) {
	t[++cnt].nxt = head[u], t[cnt].ver = v, t[cnt].val = w, head[u] = cnt;
}
inline void Init() {
	for (register int i = 1; i <= n; i++)
		fa[i] = i;
}
inline void kruskal() {
	sort(e + 1, e + m + 1, cmp);
	for (register int i = 1; i <= m; i++) {
		int eu = find(e[i].u), ev = find(e[i].v);
		if (eu == ev) continue;
		add(e[i].u, e[i].v, e[i].w), add(e[i].v, e[i].u, e[i].w);
		fa[ev] = eu;
		++tot;
		if (tot == n - 1) break;
	}
}
void DFS(int u, int fa, int w) {
	deep[u] = deep[fa] + 1;
	f[u][0] = fa, h[u][0] = w;
	for (register int i = 1; i <= 20; i++)
		f[u][i] = f[f[u][i - 1]][i - 1],
		h[u][i] = max(h[u][i - 1], h[f[u][i - 1]][i - 1]);
	for (register int i = head[u]; i; i = t[i].nxt) {
		int v = t[i].ver;
		if (v != fa)
			DFS(v, u, t[i].val);
	}
}
inline int LCA(int u, int v) {
	int ans = 0;
	if (deep[u] < deep[v]) swap(u, v);
	for (register int i = 20; i >= 0; i--)
		if (deep[f[u][i]] >= deep[v]) {
			ans = max(ans, h[u][i]);
			u = f[u][i];
		}
	if (u == v) return ans;
	for (register int i = 20; i >= 0; i--)
		if (f[u][i] != f[v][i]) {
			ans = max(ans, h[u][i]), ans = max(ans, h[v][i]);
			u = f[u][i], v = f[v][i];
		}
	ans = max(ans, h[u][0]), ans = max(ans, h[v][0]);
	return ans;
}
int main() {
	//freopen("a.in", "r", stdin);
	//freopen("1.out", "w", stdout);
	n = read(), m = read();
	for (register int i = 1; i <= m; i++)
		e[i].u = read(), e[i].v = read(), e[i].w = read();
	Init();
	kruskal();
	for (register int i = 1; i <= n; i++)
		if (!deep[i])
			DFS(i, 0, 0);
	q = read();
	for (register int i = 1; i <= q; i++) {
		u = read(), v = read();
		if (find(u) != find(v))
			printf("impossible
");
		else
			printf("%d
", LCA(u, v));
	}
	return 0;
}