zoukankan      html  css  js  c++  java
  • The Largest Clique (uva11324)

    求图的缩点

    #include<bits/stdc++.h>
    using namespace std;
    
    struct my{
       int v,next;
    };
    
    const int maxn=10000+10;
    const int maxn1=100000+10;
    my bian[maxn1];
    my bian2[maxn1];
    stack<int>s;
    int sccno[maxn];
    int adj[maxn];
    int adj2[maxn];
    int dp[maxn];
    int w[maxn];
    int fa;
    int fa2;
    int m,n,dfsnum;
    int pre[maxn];
    int lowlink[maxn];
    
    void myinsert(int u,int v){
         bian[++fa].v=v;
         bian[fa].next=adj[u];
         adj[u]=fa;
    }
    
    void myinsert2(int u,int v){
         bian2[++fa2].v=v;
         bian2[fa2].next=adj2[u];
         adj2[u]=fa2;
    }
    
    void tarjan(int u){
         pre[u]=lowlink[u]=++dfsnum;
         s.push(u);
         for (int i=adj[u];i!=-1;i=bian[i].next){
            int v=bian[i].v;
            if(!pre[v]){
                tarjan(v);
                lowlink[u]=min(lowlink[v],lowlink[u]);
            }
            else if(!sccno[v]){
                lowlink[u]=min(lowlink[u],pre[v]);
            }
         }
         if(lowlink[u]==pre[u]){
            dfsnum++;
            for (;;){
                int e=s.top();
                s.pop();
                sccno[e]=dfsnum;
                if(e==u) break;
            }
         }
    }
    void init(){
      memset(bian,-1,sizeof(bian));
      memset(bian2,-1,sizeof(bian2));
      memset(adj2,-1,sizeof(adj2));
      memset(adj,-1,sizeof(adj));
      memset(lowlink,0,sizeof(lowlink));
      memset(pre,0,sizeof(pre));
      memset(sccno,0,sizeof(sccno));
      memset(dp,-1,sizeof(dp));
      fa=fa2=0;
      memset(w,0,sizeof(w));
      dfsnum=0;
      while(!s.empty()) s.pop();
    }
    
    void build(){
        for (int i=1;i<=n;i++){
            w[sccno[i]]++;
        }
        for (int i=1;i<=n;i++){
            for (int j=adj[i];j!=-1;j=bian[j].next){
                int v=bian[j].v;
                if (sccno[i] != sccno[v]){
                    myinsert2(sccno[i],sccno[v]);
                }
            }
        }
    }
    
    int dfs(int u){
        if (dp[u] != -1) return dp[u];
        dp[u] = w[u];
        for (int i = adj2[u];i!=-1;i=bian2[i].next){
            int v = bian2[i].v;
            dp[u] = max(dp[u],dfs(v) + w[u]);
        }
        return dp[u];
    }
    
    int main(){
        int t;
        scanf("%d",&t);
        while(t--){
            int u,v;
            init();
            scanf("%d%d",&n,&m);
            for (int i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            myinsert(u,v);
           }
           for (int i=1;i<=n;i++){
            if(!pre[i])
                tarjan(i);
           }
           build();
           int ans=0;
           for (int i=1;i<=n;i++)
                ans=max(ans,dfs(sccno[i]));
            printf("%d
    ", ans);
        }
    return 0;
    }
  • 相关阅读:
    Linux内核源码分析方法
    OVS处理upcall流程分析
    Linux内核源码目录结构分析
    理解OpenStack中的OpenvSwitch的几个要点
    OVS源码connmgr_run分析
    ovs-appctl 命令合集
    云计算底层技术-使用openvswitch
    OVS架构
    Open vSwitch Datapath浅析
    Openvswitch原理与代码分析(4):网络包的处理过程
  • 原文地址:https://www.cnblogs.com/lmjer/p/8227692.html
Copyright © 2011-2022 走看看