Description

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。
Input

输入n<=100000 m<=500000及m条边
Output

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。
Sample Input
5 5

1 2

2 3

1 3

3 4

4 5
Sample Output
8

8

16

14

8

解题方法：

不懂Tarjan求割点的可以看这个，Tarjan求割点

话说为啥cout输出让我无限re啊，囧。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100005;
const int maxm = 1000005;
int n, m, head[maxn], edgecnt, ind, low[maxn], dfn[maxn], siz[maxn];
LL ans[maxn];
struct edge{int v, nxt; } E[maxm];
void init(){
    memset(head, -1, sizeof(head)); edgecnt = ind = 0;
}
void addedge(int u, int v){
    E[edgecnt].v = v, E[edgecnt].nxt = head[u], head[u] = edgecnt++;
}
void tarjan(int x)
{
    int t = 0;
    siz[x] = 1;
    low[x] = dfn[x] = ++ind;
    for(int i = head[x]; ~i; i = E[i].nxt){
        int v = E[i].v;
        if(dfn[v]){
            low[x] = min(low[x], dfn[v]);
        }
        else{
            tarjan(v);
            siz[x] += siz[v];
            low[x] = min(low[x], low[v]);
            if(dfn[x] <= low[v]){
                ans[x] += (LL) t * siz[v];
                t += siz[v];
            }
        }
    }
    ans[x] += (LL) t * (n - t - 1);
}
int main(){
    init();
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++){
        int u, v;
        scanf("%d%d", &u, &v);
        addedge(u, v);
        addedge(v, u);
    }
    tarjan(1);
    for(int i = 1; i <= n; i++){
        printf("%lld\n", (ans[i] + n - 1) * 2);
    }
    return 0;
}