UVa11324 最大团 The Largest Clique-有向图强连通分量&DP

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

 #include<iostream>
 #include<cstdio>
 #include<vector>
 #include<cstring>
 using namespace std;
 ;
 struct Edge{
     int go,next;
 };
 struct InputEdge{
     int from,to;
 };
 int T,a,b,n,m,book[maxn],bcc_count,v[maxn],end[maxn],end2[maxn],map[maxn][maxn],bcc_node[maxn],mem[maxn];
 vector<int> G[maxn],G2[maxn];
 vector<InputEdge> inputedge;
 vector<int> S;
 void init(){
     memset(v,,sizeof(v));
     memset(book,,sizeof(book));
     bcc_count=;
     //memset(end,0,sizeof(end));
     //memset(end2,0,sizeof(end2));
     S.clear();
     ;i<=n;i++){
         G[i].clear();
         G2[i].clear();
     }
     memset(map,,sizeof(map));
     inputedge.clear();
     memset(bcc_node,,sizeof(bcc_node));
     memset(mem,,sizeof(mem));
 }
 void add(int from,int to){
     //Edge e;e.go=to;e.next=end[from];G.push_back(e);end[from]=G.size()-1;
     G[from].push_back(to);
 }
 void add2(int from,int to){
     //Edge e;e.go=to;e.next=end2[from];G2.push_back(e);end2[from]=G2.size()-1;
     G2[from].push_back(to);
 }
 void dfs(int u){
     v[u]=;
     ;i<G[u].size();i++){
         //int go=G[i].go;
         int go=G[u][i];
         if(!v[go]) dfs(go);
     }
     S.push_back(u);
 }
 void dfs2(int u){
     book[u]=bcc_count;
     bcc_node[bcc_count]++;
     ;i<G2[u].size();i++){
         //int go=G2[i].go;
         int go=G2[u][i];
         if(!book[go]) dfs2(go);
     }
 }
 int dp(int x){
     if(mem[x]) return mem[x];
     int& ans=mem[x];
     ;i<=bcc_count;i++) if(i!=x&&map[i][x]) ans=max(ans,dp(i)+bcc_node[x]);
     if(!ans) ans=bcc_node[x];
     return ans;
 }
 int main()
 {
     scanf("%d",&T);
     while(T--){
         scanf("%d %d",&n,&m);
         init();
         ;i<=m;i++){
             scanf("%d %d",&a,&b);
             add(a,b);add2(b,a);
             inputedge.push_back((InputEdge){a,b});
         }
         ;i<=n;i++) if(!v[i]) dfs(i);
         ;i>=;i--) if(!book[S[i]]){
             bcc_count++;
             dfs2(S[i]);
         }
         ){
             printf("%d\n",n);
             continue;
         }
         ;i<inputedge.size();i++){
             InputEdge& e=inputedge[i];
             map[book[e.;
         }
         ;
         ;i<=bcc_count;i++)
             ans=max(ans,dp(i));
         printf("%d\n",ans);
     }
     ;
 }

用邻接表存图求BCC会出错....