uva11324 - The Largest Clique 有向图强连通分量+缩点+DP

Given a directed graph G, consider the following transformation. First, create a new graphT(G) to have the same vertex set as G. Create a directed edge betweentwo vertices u and v in T(G) if and only if there is a pathbetween u and v in G that follows the directed edges only in the forwarddirection. 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 thatfor any two vertices u and v in U, there is a directededge 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.

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 integersn and m, where 0 ≤ n ≤ 1000 is the number ofvertices of Gand 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 vbetween 1 and n which definea directed edge from u to v in G.

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

Output for sample input

4

  给出有向图，求一个节点数最大的节点集，使节点集中任意两个点u和v满足：u可以到v或者v可以到u或者可以相互到达。

  同一个强连通分量中的点要么都选，要么都不选，求出强连通分量并记录每个强连通分量的节点数，把这个强连通分量看成一个点，这个点的权值为强连通分量包含的节点数。缩点之后，把可达的点建边，这样形成了一个无环图，用记忆化搜索找一条权值最大的路径。

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<algorithm>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXN 1010
#define MAXM 2000010
#define MAXNODE 105
#define MOD 100000
#define SIGMA_SIZE 4
typedef long long LL;
using namespace std;
int T,N,M,pre[MAXN],low[MAXN],sccno[MAXN],dfs_clock,scc_cnt;
int p[MAXN],dp[MAXN],vis[MAXN][MAXN];
stack<int> S;
vector<int> G[MAXN],V[MAXN];
void dfs(int u){
    pre[u]=low[u]=++dfs_clock;
    S.push(u);
    int L=G[u].size();
    for(int i=0;i<L;i++){
        int v=G[u][i];
        if(!pre[v]){
            dfs(v);
            low[u]=min(low[u],low[v]);
        }
        else if(!sccno[v]) low[u]=min(low[u],pre[v]);
    }
    if(low[u]==pre[u]){
        scc_cnt++;
        for(;;){
            int x=S.top();
            S.pop();
            sccno[x]=scc_cnt;
            p[scc_cnt]++;
            if(x==u) break;
        }
    }
}
void find_scc(){
    dfs_clock=scc_cnt=0;
    memset(pre,0,sizeof(pre));
    memset(sccno,0,sizeof(sccno));
    memset(p,0,sizeof(p));
    for(int i=1;i<=N;i++) if(!pre[i]) dfs(i);
}
int DP(int u){
    int& ans=dp[u];
    if(ans!=-1) return ans;
    ans=p[u];
    int L=V[u].size(),MAX=0;
    for(int i=0;i<L;i++) MAX=max(MAX,DP(V[u][i]));
    ans+=MAX;
    return ans;
}
int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++) G[i].clear();
        int u,v;
        while(M--){
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
        }
        find_scc();
        for(int i=1;i<=scc_cnt;i++) V[i].clear();
        memset(vis,0,sizeof(vis));
        for(int u=1;u<=N;u++){
            int L=G[u].size();
            for(int i=0;i<L;i++){
                int v=G[u][i],m=sccno[u],n=sccno[v];
                if(m!=n&&!vis[m][n]){
                    V[m].push_back(n);
                    vis[m][n]=1;
                }
            }
        }
        int ans=0;
        memset(dp,-1,sizeof(dp));
        for(int i=1;i<=scc_cnt;i++) if(dp[i]==-1) ans=max(ans,DP(i));
        printf("%d\n",ans);
    }
    return 0;
}