We Need More Bosses CodeForces - 1000E (无向图缩点)

大意: 给定无向连通图, 定义两个点$s,t$个价值为切断一条边可以使$s,t$不连通的边数. 求最大价值.

显然只有桥会产生贡献. 先对边双连通分量缩点建树, 然后求直径即为答案.

#include <iostream>
#include <cstdio>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define pb push_back
using namespace std;

const int N = 3e5+10;
int n,m,clk,ans,cnt;
int dfn[N],low[N],dep[N],q[N],bcc[N];
vector<int> g[N],gg[N];

void dfs(int x, int fa) {
	dfn[x]=low[x]=++clk;
	q[++*q] = x;
	for (int y:g[x]) if (y!=fa) {
		if (dfn[y]) low[x]=min(low[x],dfn[y]);
		else {
			dfs(y,x);
			low[x] = min(low[x],low[y]);
		}
	}
	if (low[x]==dfn[x]) {
		++cnt;
		do bcc[q[*q]] = cnt; while (q[(*q)--]!=x);
	}
}
void dfs2(int x, int fa) {
	for (int y:gg[x]) if (y!=fa) {
		dfs2(y,x);
		ans = max(ans, dep[x]+dep[y]+1);
		dep[x] = max(dep[x],dep[y]+1);
	}
}
int main() {
	scanf("%d%d", &n, &m);
	while (m--) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0);
	REP(i,1,n) for (int j:g[i]) {
		if (bcc[i]!=bcc[j]) {
			gg[bcc[i]].pb(bcc[j]);
		}
	}
	dfs2(1,0);
	printf("%d\n", ans);
}