对于每一个点双联通分量，如果它与两个及以上割点相连，则不需要修建逃生出口，因为其一定可以通过其他割点走到其他逃生出口。如果它只与一个割点相连，则一定需要修建一个逃生出口。

#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;

const int maxn = 505;
struct edge {
    int to, next;
}e[maxn * 2];
int m, n, h[maxn], tot;
int dfn[maxn], low[maxn], Time;
int vis[maxn];
bool iscut[maxn];
long long ans1, ans2;

inline void add(int u, int v)
{
    e[++tot] = (edge) {v, h[u]};
    h[u] = tot;
} 

inline void chkmax(int &a, int b)
{
    if(a < b) a = b;
}

inline void chkmin(int &a, int b)
{
    if(a > b) a = b;
}

int root;

void tarjan(int u, int fa)
{
    dfn[u] = low[u] = ++Time;
    vis[u] = 1;
    int child = 0;
    for(int i = h[u]; i; i = e[i].next) {
      int v = e[i].to;
      if(v == fa) continue;
      if(!dfn[v]) {
        child++; tarjan(v, u);
        chkmin(low[u], low[v]);
        if(u == root && child >= 2) iscut[u] = true;
        else if(u != root && dfn[u] <= low[v]) iscut[u] = true;
      }else if(vis[v] == 1) chkmin(low[u], dfn[v]);
    }
    if(u == fa && child == 1) iscut[u] = false;
    vis[u] = 2;
}

int size, sum, now;
void dfs(int u)
{
    vis[u] = now; size++;
    for(int i = h[u]; i; i = e[i].next)
      if(!vis[e[i].to] && !iscut[e[i].to]) dfs(e[i].to);
      else if(vis[e[i].to] != now && iscut[e[i].to]) vis[e[i].to] = now, sum++;
}

int main()
{
    int i, u, v, Case = 0;
    while(scanf("%d", &m) == 1 && m) {
      memset(h, 0, sizeof(h));
      memset(dfn, 0, sizeof(dfn));
      memset(vis, 0, sizeof(vis));
      memset(iscut, 0, sizeof(iscut));
      tot = n = Time = 0;
      for(i = 1; i <= m; i++) {
        scanf("%d%d", &u, &v);
        add(u, v); add(v, u);
        chkmax(n, u); chkmax(n, v);
      }
      for(i = 1; i <= n; i++)
        if(!dfn[i]) root = i, tarjan(i, i);
      ans1 = 0; ans2 = 1;
      memset(vis, 0, sizeof(vis));
      for(i = 1; i <= n; i++)
        if(!vis[i] && !iscut[i]) {
          now++; size = sum = 0; dfs(i);
          if(sum == 1) ans1++, ans2 *= size;
        }
      if(ans1 == 0) ans1 = 2, ans2 = n * (n - 1) / 2;
      printf("Case %d: %lld %lld\n", ++Case, ans1, ans2);
    }
    return 0;
}

ps：如果这道题为非连通图则我的代码还有错误，但并没有这样的数据，所以我开开心心的AC啦