正题

题目链接:https://www.luogu.com.cn/problem/P6113

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题目大意

给出一张无向图，求最大匹配。

$1\le n\le 10^3,1\le m\le 5\times 10^4$

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解题思路

带花树的模板，我也不会讲/kel

所以看下面两篇大佬的博客吧
yyb-带花树算法学习笔记
Bill Yang-带花树学习笔记

时间复杂度好像是$O(n^3)$的

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int N=1010,M=3e5+10;
struct node{int to,next;
}a[M];
int n,m,tot,tk,ans,ls[N],dfn[N];
int fa[N],pre[N],tag[N],match[N];
queue<int> q;
void addl(int x,int y){a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;return;
}
int find(int x)
{return (fa[x]==x)?x:(fa[x]=find(fa[x]));}
int LCA(int x,int y){++tk;x=find(x);y=find(y);while(dfn[x]!=tk){dfn[x]=tk;x=find(pre[match[x]]);if(y)swap(x,y);}return x;
}
void Blossom(int x,int y,int lca){while(find(x)!=lca){pre[x]=y;y=match[x];if(tag[y]==2){tag[y]=1;q.push(y);}fa[x]=fa[y]=lca;x=pre[y];}return;
}
int Aug(int s){memset(tag,0,sizeof(tag));memset(pre,0,sizeof(pre));for(int i=1;i<=n;i++)fa[i]=i;while(!q.empty())q.pop();q.push(s);tag[s]=1;while(!q.empty()){int x=q.front();q.pop();for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(!tag[y]){tag[y]=2;pre[y]=x;if(!match[y]){for(int u=y,lst;u;u=lst)lst=match[pre[u]],match[u]=pre[u],match[pre[u]]=u;return 1;}tag[match[y]]=1;q.push(match[y]);}else if(tag[y]==1&&find(y)!=find(x)){int lca=LCA(x,y);Blossom(x,y,lca);Blossom(y,x,lca);}}}return 0;
}
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);addl(x,y);addl(y,x);}for(int i=1;i<=n;i++)if(!match[i])ans+=Aug(i);printf("%d\n",ans);for(int i=1;i<=n;i++)printf("%d ",match[i]);return 0;
}