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  • HDOJ 5409 CRB and Graph 无向图缩块


    无向图缩块后,以n所在的块为根节点,dp找每块中的最大值.

    对于每一个桥的答案为两块中的较小的最大值和较小的最大值加1

    CRB and Graph

    Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 113    Accepted Submission(s): 41


    Problem Description
    A connected, undirected graph of N vertices and M edges is given to CRB.
    A pair of vertices (uv) (u < v) is called critical for edge e if and only if u and v become disconnected by removing e.
    CRB’s task is to find a critical pair for each of M edges. Help him!
     

    Input
    There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
    The first line contains two integers NM denoting the number of vertices and the number of edges.
    Each of the next M lines contains a pair of integers a and b, denoting an undirected edge between a and b.
    1 ≤ T ≤ 12
    1 ≤ NM ≤ 105
    1 ≤ ab ≤ N
    All given graphs are connected.
    There are neither multiple edges nor self loops, i.e. the graph is simple.

     

    Output
    For each test case, output M lines, i-th of them should contain two integers u and v, denoting a critical pair (uv) for the i-th edge in the input.
    If no critical pair exists, output "0 0" for that edge.
    If multiple critical pairs exist, output the pair with largest u. If still ambiguous, output the pair with smallest v.
     

    Sample Input
    2 3 2 3 1 2 3 3 3 1 2 2 3 3 1
     

    Sample Output
    1 2 2 3 0 0 0 0 0 0
     

    Author
    KUT(DPRK)
     

    Source
     




    /* ***********************************************
    Author        :CKboss
    Created Time  :2015年08月22日 星期六 10时24分13秒
    File Name     :HDOJ5409.cpp
    ************************************************ */
    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <string>
    #include <cmath>
    #include <cstdlib>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    
    using namespace std;
    
    const int maxn=100100;
    
    typedef long long int LL;
    typedef pair<int,int> pII;
    
    struct Edge
    {
    	int from,to,next,id;
    }edge[maxn*2];
    
    int Adj[maxn],Size,n,m;
    
    void init()
    {
    	Size=0; memset(Adj,-1,sizeof(Adj));
    }
    
    void Add_Edge(int u,int v,int id)
    {
    	edge[Size].from=u;
    	edge[Size].id=id;
    	edge[Size].to=v;
    	edge[Size].next=Adj[u];
    	Adj[u]=Size++;
    }
    
    int Low[maxn],DFN[maxn],Stack[maxn],Belong[maxn];
    int Index,top,scc;
    bool Instack[maxn],vis[maxn],ve[maxn*2];
    
    void Tarjan(int u,int fa)
    {
    	int v;
    	Low[u]=DFN[u]=++Index;
    	Stack[top++]=u;
    	Instack[u]=true;
    
    	for(int i=Adj[u];~i;i=edge[i].next)
    	{
    		v=edge[i].to;
    		if(v==fa&&ve[edge[i].id]) continue;
    		ve[edge[i].id]=true;
    		if(!DFN[v])
    		{
    			Tarjan(v,u);
    			Low[u]=min(Low[u],Low[v]);
    		}
    		else
    		{
    			Low[u]=min(Low[u],DFN[v]);
    		}
    	}
    	if(Low[u]==DFN[u])	
    	{
    		scc++;
    		do
    		{
    			v=Stack[--top];
    			Belong[v]=scc;
    			Instack[v]=false;
    		}while(v!=u);
    	}
    }
    
    void scc_solve()
    {
    	memset(DFN,0,sizeof(DFN));
    	memset(Instack,0,sizeof(Instack));
    
    	Index=scc=top=0;
    	memset(ve,0,sizeof(ve));
    
    	for(int i=1;i<=n;i++)
    	{
    		if(!DFN[i]) Tarjan(i,i);
    	}
    }
    
    int value[maxn];
    vector<pII> G[maxn];
    int ans[maxn][2];
    int bian[maxn][2];
    int MX[maxn];
    
    void dfs(int u,int fa)
    {
    	MX[u]=value[u];
    	for(int i=0,sz=G[u].size();i<sz;i++)
    	{
    		int v=G[u][i].first;
    		if(v==fa) continue;
    		dfs(v,u);
    		MX[u]=max(MX[u],MX[v]);
    	}
    }
    
    int main()
    {
    	//freopen("in.txt","r",stdin);
    	//freopen("out.txt","w",stdout);
    
    	int T_T;
    	scanf("%d",&T_T);
    	while(T_T--)
    	{
    		scanf("%d%d",&n,&m);
    		init();
    		for(int i=0;i<m;i++)
    		{
    			int a,b;
    			scanf("%d%d",&a,&b);
    			bian[i][0]=a; bian[i][1]=b;
    			Add_Edge(a,b,i); Add_Edge(b,a,i);
    		}
    		scc_solve();
    
    		/***************REBUILD**********************/
    
    		memset(value,0,sizeof(value));
    		memset(ans,0,sizeof(ans));
    		int root=0;
    
    		for(int i=1;i<=n;i++)
    		{
    			G[i].clear();
    			int b=Belong[i];
    			value[b]=max(value[b],i);
    			if(value[b]==n) root=b;
    		}
    
    		//for(int i=1;i<=scc;i++) cout<<i<<" value: "<<value[i]<<endl;
    
    		for(int i=0;i<m;i++)
    		{
    			int u=Belong[bian[i][0]];
    			int v=Belong[bian[i][1]];
    			if(u==v) continue;
    			G[u].push_back(make_pair(v,i)); G[v].push_back(make_pair(u,i));
    		}
    
    		dfs(root,root);
    
    		//for(int i=1;i<=scc;i++) { cout<<i<<" mx: "<<MX[i]<<endl; }
    
    		for(int i=0;i<m;i++)
    		{
    			int u=Belong[bian[i][0]];
    			int v=Belong[bian[i][1]];
    			if(u==v) 
    			{
    				puts("0 0"); continue;
    			}
    			else
    			{
    				int mx=min(MX[u],MX[v]);
    				printf("%d %d
    ",mx,mx+1);
    			}
    		}
    	}
        
        return 0;
    }
    




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  • 原文地址:https://www.cnblogs.com/liguangsunls/p/7029325.html
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