*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;

}