*题目描述：
Byteotia城市有n个 towns m条双向roads. 每条 road 连接 两个不同的 towns ,没有重复的road. 所有towns连通。
*输入
输入n<=100000 m<=500000及m条边
*输出：
输出n个数，代表如果把第i个点去掉，将有多少对点不能互通。
*样例输入：
5 5
1 2
2 3
1 3
3 4
4 5
*样例输出：
8
8
16
14
8
*题解：
裸的找割点。答案就是每个割点的子树的大小乘上除了子树外的点，还有子树和子树之间的对数，最后还有每个点和其他n-1个点的对数。
*代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
#ifdef CT
#define debug(...) printf(__VA_ARGS__)
#define setfile()
#else
#define debug(...)
#define filename ""
#define setfile() freopen(filename".in", "r", stdin); freopen(filename".out", "w", stdout);
#endif
#define R register
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
#define dmax(_a, _b) ((_a) > (_b) ? (_a) : (_b))
#define dmin(_a, _b) ((_a) < (_b) ? (_a) : (_b))
#define cmax(_a, _b) (_a < (_b) ? _a = (_b) : 0)
#define cmin(_a, _b) (_a > (_b) ? _a = (_b) : 0)
char B[1 << 15], *S = B, *T = B;
inline int FastIn()
{
R char ch; R int cnt = 0; R bool minus = 0;
while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
ch == '-' ? minus = 1 : cnt = ch - '0';
while (ch = getc(), ch >= '0' && ch <= '9') cnt = cnt * 10 + ch - '0';
return minus ? -cnt : cnt;
}
#define maxn 100010
#define maxm 1000010
struct Edge
{
Edge *next;
int to;
}*last[maxn], e[maxm], *ecnt = e;
inline void link(R int a, R int b)
{
*++ecnt = (Edge) {last[a], b}; last[a] = ecnt;
}
int dfn[maxn], low[maxn], timer, n, m, size[maxn];
long long ans[maxn];
void dfs(R int x, R int fa)
{
dfn[x] = low[x] = ++timer;
size[x] = 1;
R int tmp = 0;
for (R Edge *iter = last[x]; iter; iter = iter -> next)
{
R int pre = iter -> to;
if (pre != fa)
{
if (!dfn[pre])
{
dfs(pre, x);
size[x] += size[pre];
cmin(low[x], low[pre]);
if (dfn[x] <= low[pre])
{
ans[x] += 1ll * tmp * size[pre];
tmp += size[pre];
}
}
else cmin(low[x], dfn[pre]);
}
}
ans[x] += 1ll * tmp * (n - 1 - tmp);
}
int main()
{
// setfile();
n = FastIn(), m = FastIn();
for (R int i = 1; i <= m; ++i)
{
R int a = FastIn(), b = FastIn();
link(a, b); link(b, a);
}
dfs(1, 0);
for (R int i= 1; i <= n; ++i) printf("%lld\n", (ans[i] + n - 1) << 1 );
return 0;
}