BZOJ1123 BLO [Tarjan][点双连通分量]

就是点双连通分量的裸题，我们记录一个siz，如果某个点可以搜到low比它的deep值大的点的话，显然是产生了一个分割后的连通块。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<map>#include<set>#include<vector>#include<queue>#define pa pair<int,int>#define inf 1000000000#define ll long long using namespace std;inline void read(int &res){    int flag=1;static char ch;    while((ch=getchar())<'0'||ch>'9')if(ch=='-')flag=-1;res=ch-48;    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-48;res*=flag;}int n,m,cnt,tot;int head[100005],size[100005],dfn[100005],low[100005];ll ans[100005];struct data{int to,next;}E[1000005];void addedge(int u,int v){    E[++tot].to=v;E[tot].next=head[u];head[u]=tot;    E[++tot].to=u;E[tot].next=head[v];head[v]=tot;}void tarjan(int x){    int t=0;    size[x]=1;    dfn[x]=low[x]=++cnt;    for(int i=head[x];i;i=E[i].next)        if(dfn[E[i].to])low[x]=min(low[x],dfn[E[i].to]);        else{            tarjan(E[i].to);            size[x]+=size[E[i].to];            low[x]=min(low[x],low[E[i].to]);            if(dfn[x]<=low[E[i].to]){                ans[x]+=(ll)t*size[E[i].to];                t+=size[E[i].to];            }        }    ans[x]+=(ll)t*(n-t-1);}int main(){    read(n),read(m);    for(int u,v,i=1;i<=m;i++)        read(u),read(v),addedge(u,v);    tarjan(1);    for(int i=1;i<=n;i++)        printf("%lld\n",(ans[i]+n-1)*2);    return 0;}