[POI2008]BLO-Blockade--Tarjan割点

Luogu 3469ACwing 365

题目分析：

• 对于一个点$u$,分此点是否为割点进行讨论：

• 若$u$不是割点，则将此点删除，其他$n-1$个点仍联通，则$u$与其他$n-1$个点不连通，由于求的是有序点对$(x,y)$,则$ans[u]=2\ast (n-1)$

• 若$u$为割点，$u$有$t$个$v$,则删除$u$之后，最多产生$t+2$个连通块：
  $u$自己为一个连通块
  $t$个$v$各自为一个连通块
  其他的节点构成一个连通块

• 对于$u$的$v$，若$dfn[u]<=low[v],sum=\sum size[v]$
  $ans[u]=size[v_1]\ast (n-size[v_1])+size[v_2]\ast (n-size[v_2])+......+size[v_t]\ast (n-size[v_t])+(n-sum-1)\ast (sum+1)$

Code:

#include <bits/stdc++.h>
using namespace std;
#define maxn 100010
#define maxm 500010

int n,m,size=0,head[maxn],f[maxn],cnt=0,dfn[maxn],low[maxn];
long long ans[maxn];
bool co[maxn];
struct edge {
	int v,nxt;
}e[maxm<<1];

inline void init_() {
	freopen("a.txt","r",stdin);
}

inline int read_() {
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}

inline void add_(int u,int v) {
	e[++size].v=v;
	e[size].nxt=head[u];
	head[u]=size;
}

inline void clean_() {
	memset(co,false,sizeof(co));
	memset(head,-1,sizeof(head));
	memset(dfn,0,sizeof(dfn));
}

void Tarjan_(int u) {
	dfn[u]=low[u]=++cnt;
	f[u]=1;ans[u]=0;
	int flag=0,sum=0;
	for(int i=head[u];~i;i=e[i].nxt) {
		int v=e[i].v;
		if(!dfn[v]) {
			Tarjan_(v);
			low[u]=min(low[u],low[v]);
			f[u]+=f[v];
			if(dfn[u]<=low[v]) {
				++flag;
				ans[u]+=(long long) f[v]*( (long long) n-f[v] );
				sum+=f[v];
				if(u!=1||flag>1) co[u]=true;
			}
		}
		else low[u]=min(low[u],dfn[v]);
	}
	if(co[u]) {
		ans[u]+=(long long) n-1;
		ans[u]+=(long long) ( (long long) n-sum-1 ) * ( (long long) sum+1 );
	}
	else {
		ans[u]=(long long) 2 * ( (long long) n-1 );
	}
}

void readda_() {
	n=read_();m=read_();
	clean_();
	int x,y;
    for(int i=1;i<=m;++i) {
		x=read_();y=read_();
		if(x==y) continue;
		add_(x,y);add_(y,x);
	}
	Tarjan_(1);
	for(int i=1;i<=n;++i) printf("%lld\n",ans[i]);
}

int main() {
	init_();
	readda_();
	return 0;
}