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  • 透彻tarjan

    tarjan 求强连通分量:

    #include<cstdio>
    #include<iostream>
    #include<cstdlib>
    #define N 1000000
    #include<vector>
    vector <int> scc;
    int sta[N],dfn[N],low[N],in[N],tar[N],tot,tp,cnt;
    void tarjan(int x)
    {
    	dfn[x]=low[x]=++tot;
    	sta[++tp]=x;
    	in[x]=1;
    	for(int i=head[x];i;i=e[i].nxt)
    	{
    		if(!dfn[e[i].to])
    		{
    			tarjan(e[i].to);
    			low[x]=min(low[x],low[e[i].to]);
    		}
    		else if(in[e[i].to])
    		{
    			low[x]=min(low[x],dfn[e[i].to]);
    		}
    	}
    	if(dfn[x]==low[x])
    	{
    		int y;
    		cnt++;
    		do{
    			y=sta[tp--];
    			in[y]=0;
    			tar[y]=cnt;
    			scc[cnt].push_back(y);
    		}while(x!=y);
    	}
    }
    

    tarjan缩点:

      拓扑排序的思想

    代码:

    #include<cstdio>
    #include<iostream>
    #include<cstdlib>
    #include<queue>
    #define N 100000
    using namespace std;
    int in[N],dfn[N],low[N],sta[N],tot,tp,cnt,nmb,head[N],nmb2;
    int n,m,p[N],h[N],tar[N],inn[N],dist[N];
    struct node{
    	int to,nxt,from;
    }e[N<<1],e2[N<<1];
    void add(int from,int to)
    {
    	e[++nmb]= (node) {to,head[from],from};
    	head[from]=nmb;
    }
    void add2(int from,int to)
    {
    	e2[++nmb2]= (node) {to,h[from],from};
    	h[from]=nmb2;
    }
    void tarjan(int x)
    {
    	dfn[x]=low[x]=++tot;
    	sta[++tp]=x;
    	in[x]=1;
    	for(int i=head[x];i;i=e[i].nxt)
    	{
    		int v=e[i].to;
    		if(!dfn[v])
    		{
    			tarjan(v);
    			low[x]=min(low[x],low[v]);
    		}
    		else if(in[v])
    		{
    			low[x]=min(low[x],dfn[v]);
    		}
    	}
    	if(dfn[x]==low[x])
    	{
    		int y;
    		while(y=sta[tp--])
    		{	
    			tar[y]=x;
    			in[y]=0;
    			if(x==y)break;
    			p[x]+=p[y];
    		}
    	}
    }
    int topo()
    {
    	queue <int> q;
    	for(int i=1;i<=n;i++)
    	if(tar[i]==i&&!inn[i])
    	{
    		q.push(i);
    		dist[i]=p[i];
    	}
    	while(q.size())
    	{
    		int x=q.front();
    		q.pop();
    		for(int i=h[x];i;i=e2[i].nxt)
    		{
    			int y=e2[i].to;
    			dist[y]=max(dist[x]+p[y],dist[y]);
    			inn[y]--;
    			if(inn[y]==0)q.push(y);
    		}
    	}
    	int ans=0;
    	for(int i=1;i<=n;i++)ans = max(ans,dist[i]);
    	return ans;
    }
    int main()
    {
    	scanf("%d%d",&n,&m);
    	for(int i=1;i<=n;i++)scanf("%d",&p[i]);
    	for(int i=1,x,y;i<=m;i++)
    	{
    		scanf("%d%d",&x,&y);
    		add(x,y);
    	}
    	for(int i=1;i<=n;i++) if(!dfn[i])tarjan(i);
    	for(int i=1;i<=m;i++)
    	{
    		int x=tar[e[i].from] , y=tar[e[i].to];
    		if(x!=y)
    		{
    			add2(x,y);
    			inn[y]++;
    		}
    	}
    	printf("%d
    ",topo());
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yelir/p/11551705.html
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