HDU 4005 The war（边双连通）

题意

​给定一张 \(n\) 个点 \(m\) 条边的无向连通图，加入一条边，使得图中权值最小的桥权值最大，如果能使图中没有桥则输出 \(-1\)。

思路

​先对原图边双缩点，然后变成了一棵树。在树上加一条边等价于使一条路径上的边都不是桥，那么原题转化为在树上删一条路径，使得最小的边最大。固定一条最小的边之后模拟即可。

代码

#include<bits/stdc++.h>

#define FOR(i, x, y) for(int i = (x), i##END = (y); i <= i##END; ++i)

#define DOR(i, x, y) for(int i = (x), i##END = (y); i >= i##END; --i)

template<typename T, typename _T> inline bool chk_min(T &x, const _T &y) {return y < x ? x = y, 1 : 0;}

template<typename T, typename _T> inline bool chk_max(T &x, const _T &y) {return x < y ? x = y, 1 : 0;}

typedef long long ll;

const int N = 10005;

const int M = 100005;

template<const int N, const int M, typename T> struct Linked_List

{

	int head[N], nxt[M], tot; T to[M];

	Linked_List() {clear();}

	T &operator [](const int x) {return to[x];}

	void clear() {memset(head, -1, sizeof(head)), tot = 0;}

	void add(int u, T v) {to[tot] = v, nxt[tot] = head[u], head[u] = tot++;}

	#define EOR(i, G, u) for(int i = G.head[u]; ~i; i = G.nxt[i])

};

struct edge {int to, cost;};

Linked_List<N, M << 1, edge> G;

Linked_List<N, N << 1, edge> T;

int dfn[N], low[N], stk[N], bel[N], dfn_idx, tp, bcc;

int miner[N], son[N];

int n, m;

void tarjan(int u, int fa_e)

{

	dfn[u] = low[u] = ++dfn_idx, stk[++tp] = u;

	EOR(i, G, u)

	{

		if(i == (fa_e ^ 1)) continue;

		int v = G[i].to;

		if(!dfn[v]) tarjan(v, i), chk_min(low[u], low[v]);

		else if(dfn[v] < dfn[u]) chk_min(low[u], dfn[v]);

	}

	if(dfn[u] == low[u])

	{

		bcc++;

		do bel[stk[tp]] = bcc; while(stk[tp--] != u);

	}

}

void dfs(int u, int f)

{

	miner[u] = 2e9, son[u] = 0;

	EOR(i, T, u)

	{

		int v = T[i].to, w = T[i].cost;

		if(v == f) continue;

		dfs(v, u);

		if(chk_min(miner[u], miner[v])) son[u] = v;

		if(chk_min(miner[u], w)) son[u] = v;

	}

}

int redfs(int u, int f)

{

	if(!son[u]) return 2e9;

	int res = redfs(son[u], u);

	EOR(i, T, u)

	{

		int v = T[i].to, w = T[i].cost;

		if(v == f || v == son[u]) continue;

		chk_min(res, miner[v]);

		chk_min(res, w);

	}

	return res;

}

int main()

{

	while(~scanf("%d%d", &n, &m))

	{

		G.tot = T.tot = 0;

		FOR(i, 1, n) G.head[i] = T.head[i] = -1;

		FOR(i, 1, n) dfn[i] = 0;

		bcc = dfn_idx = 0;

		FOR(i, 1, m)

		{

			int u, v, w;

			scanf("%d%d%d", &u, &v, &w);

			G.add(u, (edge){v, w}), G.add(v, (edge){u, w});

		}

		tarjan(1, -1);

		int s, t, mi = 2e9;

		FOR(u, 1, n) EOR(i, G, u)

		{

			int v = G[i].to, w = G[i].cost;

			if(bel[u] < bel[v])

			{

				if(chk_min(mi, w)) s = bel[u], t = bel[v];

				T.add(bel[u], (edge){bel[v], w}), T.add(bel[v], (edge){bel[u], w});

			}

		}

		dfs(s, t), dfs(t, s);

		int res = std::min(redfs(s, t), redfs(t, s));

		printf("%d\n", (res > 1e9 ? -1 : res));

	}

	return 0;

}