Bzoj1123 Blockade

题目链接：https://loj.ac/problem/10104

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日常水题，题目中已经给出了算法，写个模板即可，不会割点的这里有一篇博客：https://www.cnblogs.com/WWHHTT/p/9745499.html

难点是每个对可以互换顺序，然后删掉一个点之后它与其他所有点的对都会少1

下面给出代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
inline int rd(){int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=-1;for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';return x*f;
}
inline void write(long long x){if(x<0) putchar('-'),x=-x;if(x>9) write(x/10);putchar(x%10+'0');return ;
}
int n,m;
int head[1000006],nxt[1000006],to[1000006];
int total=0;
void add(int x,int y){total++;to[total]=y;nxt[total]=head[x];head[x]=total;return ;
}
int tot=0;
int dfn[1000006],low[1000006],book[1000006];
int q[1000006],l=0,r=0;
int size[1000006];
long long ans[1000006];
void Tarjan(int x){int sum=0;size[x]=1;dfn[x]=low[x]=++tot;for(int e=head[x];e;e=nxt[e]){if(!dfn[to[e]]){Tarjan(to[e]);low[x]=min(low[x],low[to[e]]);size[x]+=size[to[e]];if(low[to[e]]>=dfn[x]){ans[x]+=(long long)sum*(long long)size[to[e]];sum+=size[to[e]];}}else low[x]=min(low[x],dfn[to[e]]);}ans[x]+=(long long)sum*(long long)(n-sum-1);return ;
}
int main(){n=rd(),m=rd();for(int i=1;i<=m;i++){int x=rd(),y=rd();add(x,y),add(y,x);}for(int i=1;i<=n;i++) if(!dfn[i]) Tarjan(i);for(int i=1;i<=n;i++) write((ans[i]+n-1)*2),puts("");return 0;
}