如果n == 1 结果就是0 //wa一次
如果m == 0 结果就是n - 1 //wa一次
其他的情况 ：
看看割点去掉后，能有几个联通分量，也就是割点有几个儿子。 这个就是去掉个点后增加的分量个数。
#include <stdio.h>
#include <string.h>
const int N = 10001;
const int M = 20001;
struct e{
int v;
e* next;
};
e *link[N], edge[M * 2];
int m, n, son[N], rootdegree, el, b[N], bb[N], time;
bool judge[N];
int min(int a, int b){
return a > b ? b : a;
}
void dfs(int u, int root){
e *l = link[u];int i, v;bool c = false;
b[u] = bb[u] = time++;judge[u] = true;
while(l){
v = l->v;
if(!judge[v]){
dfs(v, root);
if(u == root){
rootdegree++;
if(rootdegree >= 2) son[u]++;
}
else{
bb[u] = min(bb[u], bb[v]);
if(bb[v] >= b[u])son[u]++;
}
}
else bb[u] = min(bb[u], b[v]);
l = l->next;
}
}
int main(){
freopen("2299.in", "r", stdin);
freopen("2299.out", "w", stdout);
int i, j, u, v, fl, ans;
while(scanf("%d%d", &n, &m), n || m){
for(i = 0; i < n; i++){
b[i] = bb[i] = judge[i] = 0;son[i] = 1;link[i] = NULL;
}el = rootdegree = fl = 0;
for(i = 0; i < m; i++){
scanf("%d%d", &u, &v);
edge[el].v = v;edge[el].next = link[u];link[u] = &edge[el++];
edge[el].v = u;edge[el].next = link[v];link[v] = &edge[el++];
}
if(n == 1){
puts("0");continue;
}
if(m == 0){
printf("%d\n", n - 1);continue;
}
for(i = 0; i < n; i++){
if(!judge[i]){
rootdegree = time = 0;dfs(i, i);fl++;
}
}ans = 1;
for(i = 0; i < n; i++){
if(son[i] > ans)ans = son[i];
}printf("%d\n", ans + fl - 1);
}
return 0;
}