P2860 [USACO06JAN]冗余路径Redundant Paths tarjan

题目链接

https://www.luogu.org/problemnew/show/P2860

思路

缩点，之后就成了个树一般的东西了
然后（叶子节点+1）/2就是答案，好像贪心的样子，lmc好像讲过诶

#include <iostream>
#include <cstdio>
#include <vector>
#include <bits/stdc++.h>
#define iter vector<int>::iterator
using namespace std;
const int N=1e5+7;
int read() {
   int x=0,f=1;char s=getchar();
   for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
   for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
   return x*f;
}
int n,m;
vector<int> G[N];
int dfn[N],low[N],SD,stak[N],top,vis[N],belong[N];
void tarjan(int u,int fa) {
   dfn[u]=low[u]=++SD;
   vis[u]=1;
   stak[++top]=u;
   for(iter it=G[u].begin();it!=G[u].end();++it) {
   	if(*it==fa) continue;
   	if(!dfn[*it]) {
   		tarjan(*it,u);
   		low[u]=min(low[u],low[*it]);
       } else if(vis[*it])
   		low[u]=min(low[u],dfn[*it]);
   }
   if(low[u]==dfn[u]) {
   	belong[0]++;
   	while(stak[top]!=u) {
   		belong[stak[top]]=belong[0];
   		vis[stak[top]]=0;
   		top--;
   	}
   	belong[u]=belong[0];
   	vis[u]=0;
   	top--;
   }
}
int ru[N];
map<pair<int,int>,int> hasH;
int main() {
   n=read(),m=read();
   for(int i=1;i<=m;++i) {
   	int x=read(),y=read();
   	if(hasH.count(make_pair(x,y))) continue;
   	hasH[make_pair(x,y)]=1;
   	G[x].push_back(y);
   	G[y].push_back(x);
   }
   for(int i=1;i<=n;++i)
   	if(!dfn[i])
   		tarjan(i,0);
   for(int i=1;i<=n;++i) {
   	for(iter it=G[i].begin();it!=G[i].end();++it) {
   		if(belong[i]!=belong[*it]) {
   			ru[belong[i]]++,ru[belong[*it]]++;
   		}
   	}
   }
   int ans=0;
   for(int i=1;i<=n;++i) if(ru[i]==2) ans++;
   printf("%d\n",(ans+1)/2);
   return 0;
}