【luogu P2860 [USACO06JAN]冗余路径Redundant Paths】 题解

题目链接：https://www.luogu.org/problemnew/show/P2860
考虑在无向图上缩点。
运用到边双、桥的知识。
缩点后统计度为1的点。
度为1是有一条路径，度为2是有两条路径。

#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 300000 + 10;
struct edge{
	int next, to, from;
}e[maxn<<2];
int head[maxn], cnt;
bool vis[maxn];
int n, m, u[maxn], v[maxn], d[maxn], ans;
stack<int> s; 
int dfn[maxn], low[maxn], tim, color[maxn], num;
void add(int u, int v)
{
	e[++cnt].from = u;
	e[cnt].next = head[u];
	e[cnt].to = v;
	head[u] = cnt;
}
void tarjan(int x)
{
	dfn[x] = low[x] = ++tim;
	for(int i = head[x]; i != -1; i = e[i].next)
	if(!vis[i])
	{
		int v = e[i].to;
		if(dfn[v]) low[x] = min(low[x], dfn[v]);
		else 
		{
			s.push(v);
			vis[i] = vis[(i&1)?i+1:i-1] = 1;
			tarjan(v);
			vis[i] = vis[(i&1)?i+1:i-1] = 0;
			low[x] = min(low[x], low[v]);
		}
	}
	if(dfn[x] == low[x])
	{
		num++;
		while(!s.empty())
		{
			color[s.top()] = num;
			if(s.top() == x) {
				s.pop();
				break;
			}
			s.pop();
		}
	}
}
int main()
{
	memset(head, -1, sizeof(head));
	scanf("%d%d",&n,&m);
	for(int i = 1; i <= m; i++)
	{
		scanf("%d%d",&u[i],&v[i]);
		add(u[i],v[i]); add(v[i],u[i]);
	}
	for(int i = 1; i <= n; i++)
		if(!dfn[i]) num = 0, s.push(i), tarjan(i);
	for(int i = 1; i <= m; i++)
	{
		if(color[u[i]] != color[v[i]])
		d[color[u[i]]]++, d[color[v[i]]]++;
	}
	for(int i = 1; i <= n; i++) if(d[i] == 1) ans++;
	printf("%d",(ans+1)/2);
	return 0;
}