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  • hdu 2586 How far away?(LCA模板题+离线tarjan算法)

    How far away ?

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 25408    Accepted Submission(s): 10111


    Problem Description
    There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
     
    Input
    First line is a single integer T(T<=10), indicating the number of test cases.
      For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
      Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
     
    Output
    For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
     
    Sample Input
    2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
     
    Sample Output
    10 25 100 100
     
     
    题目大意:
    给你一棵树,求任两点的距离。
     
    这是LCA模板题。
    求出两点的公共祖先ancst,然后dis[i,j]=depth[i]+depth[j]-depth[ancst]*2;其中depth是节点在有根树中的深度。
    离线tarjan算法。。ctrlc ctrlv一段大佬的解释吧。(来自zhouzhendong的cnblogs)
     

    LCA_Tarjan

      TarjanTarjan 算法求 LCA 的时间复杂度为 O(n+q)O(n+q) ,是一种离线算法,要用到并查集。(注:这里的复杂度其实应该不是 O(n+q)O(n+q) ,还需要考虑并查集操作的复杂度 ,但是由于在多数情况下,路径压缩并查集的单次操作复杂度可以看做 O(1)O(1),所以写成了 O(n+q)O(n+q) 。)

      TarjanTarjan 算法基于 dfs ,在 dfs 的过程中,对于每个节点位置的询问做出相应的回答。

      dfs 的过程中,当一棵子树被搜索完成之后,就把他和他的父亲合并成同一集合;在搜索当前子树节点的询问时,如果该询问的另一个节点已经被访问过,那么该编号的询问是被标记了的,于是直接输出当前状态下,另一个节点所在的并查集的祖先;如果另一个节点还没有被访问过,那么就做下标记,继续 dfs 。

      当然,暂时还没那么容易弄懂,所以建议结合下面的例子和标算来看看。

    (下面的集合合并都用并查集实现)

      比如:8114138−1−14−13 ,此时已经完成了对子树 11 的子树 1414 的 dfsdfs 与合并( 1414 子树的集合与 11 所代表的集合合并),如果存在询问 (13,14)(13,14) ,则其 LCA 即 getfather(14)getfather(14) ,即 11 ;如果还存在由节点 1313 与 已经完成搜索的子树中的 节点的询问,那么处理完。然后合并子树 1313 的集合与其父亲 11 当前的集合,回溯到子树 11 ,并深搜完所有 11 的其他未被搜索过的儿子,并完成子树 11 中所有节点的合并,再往上回溯,对节点 11 进行类似的操作即可。

     
    大意就是,后根dfs一棵树,如果子树遍历完了,就把它整个加入子树根节点的父亲节点的并查集中。
    遇到询问,就判断它的另一点是否已经遍历过了,如果是,就getfather。还是自己画画图,应该不难懂吧。
     
    第一次写,比较粗糙,附AC代码:
    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <string>
    #include <cmath>
    #include <queue>
    #include <deque>
    #include <stack>
    #include <map>
    #include <set>
    typedef long long LL;
    const int MAX_N=40000;
    const int MAX_M=200;
    const int INF=1000000000;
    
    struct tedge
    {
        int to,w,next;
    };
    tedge edge[MAX_N*2+5];
    int head1[MAX_N+5],cnt1;
    
    void addedge(int a,int b,int c)
    {
        edge[cnt1]=(tedge){b,c,head1[a]};head1[a]=cnt1++;
        edge[cnt1]=(tedge){a,c,head1[b]};head1[b]=cnt1++;
    }
    
    struct tquery
    {
        int to,next;
        int index;
    };
    tquery query[MAX_M*2+5];
    int head2[MAX_N+5],cnt2;
    
    void addquery(int a,int b,int i)
    {
        query[cnt2]=(tquery){b,head2[a],i};head2[a]=cnt2++;
        query[cnt2]=(tquery){a,head2[b],i};head2[b]=cnt2++;
    }
    
    int fa[MAX_N];
    
    int getf(int x)
    {
        if(fa[x]==x)
            return x;
        else
            return fa[x]=getf(fa[x]);
    }
    
    int vis[MAX_N+5];
    int depth[MAX_N+5];
    int ans[MAX_M+5];
    
    void LCA(int x,int pa,int dis)
    {
        for(int i=head1[x];i!=-1;i=edge[i].next)
        {
            int l=edge[i].to;
            if(l!=pa)
            {
                LCA(l,x,dis+edge[i].w);
            }
        }
        depth[x]=dis;
        for(int i=head2[x];i!=-1;i=query[i].next)
        {
            int l=query[i].to;
            if(vis[l])
            {
                int ancst=getf(l);
                ans[query[i].index]=depth[l]+depth[x]-depth[ancst]*2;
                //printf("%d %d %d
    ",l,x,ancst);
            }
        }
        fa[x]=pa;vis[x]=1;
    }
    
    void init()
    {
        memset(head1,-1,sizeof(head1));cnt1=0;
        memset(head2,-1,sizeof(head2));cnt2=0;
        memset(vis,0,sizeof(vis));
    }
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            init();
            int n,m;
            scanf("%d%d",&n,&m);
            for(int i=1,a,b,c;i<=n-1;i++)
            {
                scanf("%d%d%d",&a,&b,&c);
                addedge(a,b,c);
            }
            for(int i=1,a,b;i<=m;i++)
            {
                scanf("%d%d",&a,&b);
                addquery(a,b,i);
            }
            for(int i=1;i<=n;i++)
                fa[i]=i;
            LCA(1,-1,0);
            for(int i=1;i<=m;i++)
                printf("%d
    ",ans[i]);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/acboyty/p/10014039.html
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