|
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
|
using namespace std;
|
|
|
|
|
|
|
|
|
|
|
|
const int N = 100010, M = 300010;
|
|
|
|
|
|
|
|
|
|
|
|
// 链式前向星
|
|
|
|
|
|
int e[M], h[N], idx, w[M], ne[M];
|
|
|
|
|
|
void add(int a, int b, int c = 0) {
|
|
|
|
|
|
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
// 点双连通分量
|
|
|
|
|
|
int dfn[N], low[N], stk[N], ts, top, root;
|
|
|
|
|
|
vector<int> bcc[N];
|
|
|
|
|
|
int bcnt;
|
|
|
|
|
|
void tarjan(int u, int fa) {
|
|
|
|
|
|
int cnt = 0; // 去掉u之后还剩多少个连通块
|
|
|
|
|
|
|
|
|
|
|
|
low[u] = dfn[u] = ++ts;
|
|
|
|
|
|
stk[++top] = u;
|
|
|
|
|
|
int son = 0;
|
|
|
|
|
|
for (int i = h[u]; ~i; i = ne[i]) {
|
|
|
|
|
|
int v = e[i];
|
|
|
|
|
|
if (v == fa) continue;
|
|
|
|
|
|
if (!dfn[v]) {
|
|
|
|
|
|
son++;
|
|
|
|
|
|
tarjan(v, u);
|
|
|
|
|
|
low[u] = min(low[u], low[v]);
|
|
|
|
|
|
|
|
|
|
|
|
if (low[v] >= dfn[u]) {
|
|
|
|
|
|
cnt++; // u是割点，把u删除掉，所有的v就都孤立了，多出来1个孤立块v
|
|
|
|
|
|
|
|
|
|
|
|
int x;
|
|
|
|
|
|
bcnt++;
|
|
|
|
|
|
do {
|
|
|
|
|
|
x = stk[top--];
|
|
|
|
|
|
bcc[bcnt].push_back(x);
|
|
|
|
|
|
} while (x != v); // 将子树出栈
|
|
|
|
|
|
bcc[bcnt].push_back(u); // 把割点/树根也丢到点双里
|
|
|
|
|
|
}
|
|
|
|
|
|
}
|
|
|
|
|
|
low[u] = min(low[u], dfn[v]);
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
if (fa == -1 && son == 0) bcc[++bcnt].push_back(u); // 特判独立点,单独成点双,本题可以不用管这个
|
|
|
|
|
|
if (u != root) cnt++; // u不是根节点的话，除了自己的子树，还有他的父节点也断开了连接
|
|
|
|
|
|
ans = max(ans, cnt); // 不断更新生成的孤立块最大数量
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
int ans; // 在处理每个连通块时，尝试删除每一个割点，记录可以生成的孤立块的数量最大值
|
|
|
|
|
|
|
|
|
|
|
|
int main() {
|
|
|
|
|
|
int n, m;
|
|
|
|
|
|
while (scanf("%d %d", &n, &m), n || m) {
|
|
|
|
|
|
memset(dfn, 0, sizeof dfn);
|
|
|
|
|
|
memset(low, 0, sizeof low);
|
|
|
|
|
|
memset(stk, 0, sizeof stk);
|
|
|
|
|
|
memset(h, -1, sizeof h);
|
|
|
|
|
|
idx = ts = top = 0;
|
|
|
|
|
|
|
|
|
|
|
|
while (m--) {
|
|
|
|
|
|
int a, b;
|
|
|
|
|
|
scanf("%d %d", &a, &b);
|
|
|
|
|
|
a++, b++; // 本题点号从0开始，+1后平移到1开始 0≤a,b<n
|
|
|
|
|
|
add(a, b), add(b, a);
|
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
|
|
// 连通块的最大(多)数量
|
|
|
|
|
|
ans = 0;
|
|
|
|
|
|
|
|
|
|
|
|
int cnt = 0; // 原图中块数量(互相孤立的有多少个块)
|
|
|
|
|
|
for (root = 1; root <= n; root++)
|
|
|
|
|
|
if (!dfn[root]) {
|
|
|
|
|
|
tarjan(root, -1);
|
|
|
|
|
|
cnt++; // 记录有多少个独立的块
|
|
|
|
|
|
}
|
|
|
|
|
|
printf("%d\n", cnt + ans - 1);
|
|
|
|
|
|
}
|
|
|
|
|
|
return 0;
|
|
|
|
|
|
}
|
