BZOJ4243: 交朋友

版权声明：蒟蒻Blog随意转载 https://blog.csdn.net/a1799342217/article/details/82832536

并查集 BFS

BZOJ题目传送门

如果一个联通块里有一条双向边，这个联通块就构成了一个完全子图。

把原图中所有能开的会议都开掉，那么剩下的所有边都可以合并两个联通块。把这些点扔进队列里BFS就好了

代码：

#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100005
#define F inline
using namespace std;
struct edge{ int nxt,to; }ed[N<<1];
int n,m,l,r,k,h[N],fa[N],q[N],out[N],w[N];
bool f[N]; long long ans;
F char readc(){
	static char buf[100000],*l=buf,*r=buf;
	if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
	return l==r?EOF:*l++;
}
F int _read(){
	int x=0; char ch=readc();
	while (!isdigit(ch)) ch=readc();
	while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
	return x;
}
#define add(x,y) ed[++k]=(edge){h[x],y},h[x]=k
int findfa(int x){ return x==fa[x]?x:fa[x]=findfa(fa[x]); }
int main(){
	freopen("1.in","r",stdin);
	freopen("1.out","w",stdout);
	n=_read(),m=_read();
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1,x,y;i<=m;i++) x=_read(),y=_read(),out[x]++,add(x,y);
	for (int x=1;x<=n;x++)
		for (int i=h[x],v;i;i=ed[i].nxt)
		for (int j=h[v=ed[i].to];j;j=ed[j].nxt)
			if (ed[j].to==x){ fa[findfa(x)]=findfa(v); break; }
	for (int x=1;x<=n;x++)
		for (int i=h[x],lst=0;i;lst=ed[i].to,i=ed[i].nxt)
			if (lst) fa[findfa(lst)]=findfa(ed[i].to);
	for (int i=1;i<=n;i++) w[findfa(i)]++;
	for (int i=1;i<=n;i++) if (w[fa[i]]>1) f[q[++r]=i]=true;
	while (l<r){
		int x=q[++l];
		for (int i=h[x],v;i;i=ed[i].nxt){
			fa[findfa(v=ed[i].to)]=findfa(x);
			if (!f[v]) f[q[++r]=v]=true;
		}
	}
	for (int i=1;i<=n;i++) w[i]=0;
	for (int i=1;i<=n;i++) w[findfa(i)]++;
	for (int i=1;i<=n;i++)
		if (fa[i]==i) ans+=w[i]>1?1ll*w[i]*(w[i]-1):out[i];
	return printf("%lld\n",ans),0;
}