P2860 [USACO06JAN]冗余路径Redundant Paths

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
cpp #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; }

标签：Paths,vis,int,top,belong,Redundant,P2860,low,stak
来源： https://www.cnblogs.com/dsrdsr/p/10393850.html