题意：求一个图中删掉一个结点所得到的连通块数量最大。
思路：设一个结点u，如果u是根结点（图可能是不连通的，所以可能有多个根结点）则删掉u所增加的连通块数为 SUM(u的子结点) - 1；
如果u是非根结点，v是u的子结点，则删掉u所增加的连通块数为 SUM(u->v为割边， 既u是到v的割点)；
渣代码：
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1 #include <iostream>
2 #include <cstring>
3 #include <cstdio>
4 #include <stack>
5
6 using namespace std;
7
8 const int N = 10005;
9 const int M = 1000005;
10
11 struct node {
12 int to;
13 int next;
14 } g[M];
15
16 int head[N], num[N];
17 int dfn[N], low[N];
18 int t, ind, Rcnt, ans;
19 bool vis[N];
20
21 void init() {
22 memset(head, 0, sizeof(head));
23 memset(num, 0, sizeof(num));
24 memset(dfn, 0, sizeof(dfn));
25 memset(low, 0, sizeof(low));
26 memset(vis, 0, sizeof(vis));
27
28 t = 1; ind = Rcnt = 0;
29 }
30
31 void add(int u, int v) {
32 g[t].to = v; g[t].next = head[u]; head[u] = t++;
33 }
34
35 void tarjan(int u, int r) {
36 int v, i;
37 dfn[u] = low[u] = ++ind;
38 for(i = head[u]; i; i = g[i].next) {
39 v = g[i].to;
40 if(!dfn[v]) {
41 tarjan(v, r);
42 if(u == r) Rcnt ++;
43 else {
44 low[u] = min(low[u], low[v]);
45 if(low[v] >= dfn[u]) num[u]++;
46 }
47
48 } else {
49 low[u] = min(low[u], dfn[v]);
50 }
51 }
52 ans = max(ans, num[u]);
53 }
54
55 int main() {
56 //freopen("data.in", "r", stdin);
57
58 int n, m, u, v;
59 int i, sum;
60 while(scanf("%d%d", &n, &m), n || m) {
61 if(m == 0) { printf("%d\n", n-1); continue; }
62 init();
63 for(i = 1; i <= m; i++) {
64 scanf("%d%d", &u, &v);
65 add(u, v);
66 add(v, u);
67 }
68 sum = 0, ans = 0;
69 for(i = 0; i < n; i++) {
70 if(!dfn[i]) {
71 Rcnt = 0;
72 tarjan(i, i);
73 num[i] = Rcnt-1;
74 ans = max(num[i], ans);
75 sum++;
76 }
77 }
78 printf("%d\n", ans + sum);
79 }
80 return 0;
81 }