P4180 严格次小生成树[BJWC2010] Kruskal，倍增

题目链接\(Click\) \(Here\)。

题意就是要求一个图的严格次小生成树。以前被题面吓到了没敢做，写了一下发现并不难。

既然要考虑次小我们就先考虑最小。可以感性理解到一定有一种次小生成树，可以由最小生成树删一条边再加一条边得到。我们枚举加上去的这一条边，加上去以后原\(mst\)会成为一个基环树，想让它次小就在这个环里找一条最长的边（不包含新加进去的）删掉就好。放在树上来讲，就是找到\(u\)到\(v\)路径上的最大值。这样我们就有了非严格次小生成树。

严格要怎么处理？我们需要排除新加上的边和\(u\)到\(v\)路径最长边相等的情况。仔细思考会发现可以再倍增维护一个次大长度，如果最大长度严格小于新加上边的长度就选用最大，否则就用次大。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 100000 + 5;
const int M = 300000 + 5;
const int INF = 0x7fffffff;
const LL INFF = 0x7fffffffffffffff;

struct Graph {

	int cnt, head[N];

	struct Edge {int nxt, to, w;}e[N << 1];

	void clear () {
		cnt = -1;
		memset (head, -1, sizeof (head));
	}

	void add_len (int u, int v, int w) {
		e[++cnt] = (Edge) {head[u], v, w}; head[u] = cnt;
		e[++cnt] = (Edge) {head[v], u, w}; head[v] = cnt;
	}

	int deep[N], pre[N][20], w1st[N][20], w2nd[N][20];

	void dfs (int u, int fa) {
		deep[u] = deep[fa] + 1;
		for (int i = head[u]; ~i; i = e[i].nxt) {
			int v = e[i].to;
			if (v != fa) {
				dfs (v, u);
				pre[v][0] = u;
				w2nd[v][0] = -INF;
				w1st[v][0] = e[i].w;
			}
		}
	}

	void build (int u) {
		for (int i = 1; (1 << i) <= deep[u]; ++i) {
			pre[u][i] = pre[pre[u][i - 1]][i - 1];
			w1st[u][i] = max (w1st[u][i - 1], w1st[pre[u][i - 1]][i - 1]);
			w2nd[u][i] = max (w2nd[u][i - 1], w2nd[pre[u][i - 1]][i - 1]);
			if (w1st[u][i - 1] < w1st[pre[u][i - 1]][i - 1]) {
				w2nd[u][i] = max (w2nd[u][i], w1st[u][i - 1]);
			}
			if (w1st[u][i - 1] > w1st[pre[u][i - 1]][i - 1]) {
				w2nd[u][i] = max (w2nd[u][i], w1st[pre[u][i - 1]][i - 1]);
			}
		}
		for (int i = head[u]; ~i; i = e[i].nxt) {
			if (e[i].to != pre[u][0]) build (e[i].to);
		}
	}

	int lca (int u, int v) {
		if (deep[u] < deep[v]) swap (u, v);
		for (int i = 19; i >= 0; --i) {
			if (deep[u] - deep[v] >= (1 << i)) {
				u = pre[u][i];
			}
		}
		if (u == v) return u;
		for (int i = 19; i >= 0; --i) {
			if (pre[u][i] != pre[v][i]) {
				u = pre[u][i];
				v = pre[v][i];
			}
		}
		return pre[u][0];
	}

	int query (int u, int v, int w) {
		//求u到v之间严格小于w的最大的值
		int ans = -INF;
		for (int i = 19; i >= 0; --i) {
			if (deep[v] - deep[u] >= (1 << i)) {
				if (w1st[v][i] < w) {
					ans = max (ans, w1st[v][i]);
				} else {
					ans = max (ans, w2nd[v][i]);
				}
				v = pre[v][i];
			}
		}
		return ans;
	}

}G;	

struct Len {
	int u, v, w;

	bool operator < (Len rhs) const {
		return w < rhs.w;
	}

	Len () {}
	Len (int u, int v, int w) : u(u), v(v), w(w) {}
}arr[M];

int n, m, Set[N], in_mst[M];

int find (int x) {
	return x == Set[x] ? x : (Set[x] = find (Set[x]));
}

int main () {
	G.clear ();
	cin >> n >> m;
	for (int i = 0; i < m; ++i) {
		static int u, v, w;
		cin >> u >> v >> w;
		arr[i] = Len (u, v, w);
	}
	sort (arr, arr + m);
	for (int i = 0; i <= n; ++i) Set[i] = i;
	LL mstw = 0;
	for (int i = 0; i < m; ++i) {
		int u = arr[i].u;
		int v = arr[i].v;
		int w = arr[i].w;
		if (find (u) != find (v)) {
			mstw += w;
			in_mst[i] = true;
			G.add_len (u, v, w);
			Set[find (u)] = find (v);
		}
	}
	G.w1st[1][0] = -INF;
	G.dfs (1, 0);
	G.build (1);
	LL ans = INFF;
	for (int i = 0; i < m; ++i) {
		if (!in_mst[i]) {
			int u = arr[i].u;
			int v = arr[i].v;
			int w = arr[i].w;
			int _lca = G.lca (u, v);
			int maxu = G.query (_lca, u, w);
			int maxv = G.query (_lca, v, w);
			ans = min (ans, mstw + w - max (maxu, maxv));
		}
	}
	cout << ans << endl;
}