https://vjudge.net/problem/UVA-11324
给定一张有向图G，求一个节点数目最大的节点集，使得该集合中的任意两个节点u和v满足：要么u可以到达v，要么v可以到达u（u,v相互可达也算满足），要求输出最大的节点数
Given a directed graph G, consider the following transformation. First, create a new graph T(G) to have the same vertex set as G. Create a directed edge between two vertices u and v in T(G) if and only if there is a path between u and v in G that follows the directed edges only in the forward direction. This graph T(G) is often called the transitive closure of G. We define a clique in a directed graph as a set of vertices U such that for any two vertices u and v in U, there is a directed edge either from u to v or from v to u (or both). The size of a clique is the number of vertices in the clique.
Input The number of cases is given on the first line of input. Each test case describes a graph G. It begins with a line of two integers n and m, where 0 ≤ n ≤ 1000 is the number of vertices of G and 0 ≤ m ≤ 50, 000 is the number of directed edges of G. The vertices of G are numbered from 1 to n. The following m lines contain two distinct integers u and v between 1 and n which define a directed edge from u to v in G.
Output For each test case, output a single integer that is the size of the largest clique in T(G).
Sample Input 1 5 5 1 2 2 3 3 1 4 1 5 2
Sample Output 4
1 #include<iostream>
2 #include<cstdio>
3 #include<vector>
4 #include<cstring>
5 using namespace std;
6 const int maxn=1010;
7 struct Edge{
8 int go,next;
9 };
10 struct InputEdge{
11 int from,to;
12 };
13 int T,a,b,n,m,book[maxn],bcc_count,v[maxn],end[maxn],end2[maxn],map[maxn][maxn],bcc_node[maxn],mem[maxn];
14 vector<int> G[maxn],G2[maxn];
15 vector<InputEdge> inputedge;
16 vector<int> S;
17 void init(){
18 memset(v,0,sizeof(v));
19 memset(book,0,sizeof(book));
20 bcc_count=0;
21 //memset(end,0,sizeof(end));
22 //memset(end2,0,sizeof(end2));
23 S.clear();
24 for(int i=1;i<=n;i++){
25 G[i].clear();
26 G2[i].clear();
27 }
28 memset(map,0,sizeof(map));
29 inputedge.clear();
30 memset(bcc_node,0,sizeof(bcc_node));
31 memset(mem,0,sizeof(mem));
32 }
33 void add(int from,int to){
34 //Edge e;e.go=to;e.next=end[from];G.push_back(e);end[from]=G.size()-1;
35 G[from].push_back(to);
36 }
37 void add2(int from,int to){
38 //Edge e;e.go=to;e.next=end2[from];G2.push_back(e);end2[from]=G2.size()-1;
39 G2[from].push_back(to);
40 }
41 void dfs(int u){
42 v[u]=1;
43 for(/*int i=end[u];i;i=G[i].next*/int i=0;i<G[u].size();i++){
44 //int go=G[i].go;
45 int go=G[u][i];
46 if(!v[go]) dfs(go);
47 }
48 S.push_back(u);
49 }
50 void dfs2(int u){
51 book[u]=bcc_count;
52 bcc_node[bcc_count]++;
53 for(/*int i=end2[u];i;i=G2[i].next*/int i=0;i<G2[u].size();i++){
54 //int go=G2[i].go;
55 int go=G2[u][i];
56 if(!book[go]) dfs2(go);
57 }
58 }
59 int dp(int x){
60 if(mem[x]) return mem[x];
61 int& ans=mem[x];
62 for(int i=1;i<=bcc_count;i++) if(i!=x&&map[i][x]) ans=max(ans,dp(i)+bcc_node[x]);
63 if(!ans) ans=bcc_node[x];
64 return ans;
65 }
66 int main()
67 {
68 scanf("%d",&T);
69 while(T--){
70 scanf("%d %d",&n,&m);
71 init();
72 for(int i=1;i<=m;i++){
73 scanf("%d %d",&a,&b);
74 add(a,b);add2(b,a);
75 inputedge.push_back((InputEdge){a,b});
76 }
77 for(int i=1;i<=n;i++) if(!v[i]) dfs(i);
78 for(int i=n-1;i>=0;i--) if(!book[S[i]]){
79 bcc_count++;
80 dfs2(S[i]);
81 }
82 if(bcc_count==1){
83 printf("%d\n",n);
84 continue;
85 }
86 for(int i=0;i<inputedge.size();i++){
87 InputEdge& e=inputedge[i];
88 map[book[e.from]][book[e.to]]=1;
89 }
90 int ans=0;
91 for(int i=1;i<=bcc_count;i++)
92 ans=max(ans,dp(i));
93 printf("%d\n",ans);
94 }
95 return 0;
96 }
用邻接表存图求BCC会出错....