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  • hdu 2586

    LCA模板题

    How far away ?

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 9094    Accepted Submission(s): 3168


    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
     
    Source
     
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     题意:给一个无根树,有q个询问,每个询问两个点,问两点的距离。求出  lca = LCA(X,Y) , 然后  dir[x] + dir[y] - 2 * dir[lca]

    dir[u]表示点u到树根的距离

    下面两份代码都可以通过HDU的C++和G++,都不存在爆栈问题,网上很多人说会爆栈,加了申请系统栈语句,其实不用,而且好想比赛中不允许使用的

    Tarjan算法跑得更快些,C++ 15ms,  G++ 50ms  左右, RMQ大概60ms

     在线算法:LCA转化为RMQ

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cmath>
    using namespace std;
    //#pragma comment(linker, "/STACK:102400000,102400000") //不需要申请系统栈
    const int N = 40010;
    const int M = 25;
    
    int _pow[M];
    int tot,head[N],ver[2*N],R[2*N],first[N],dir[N];
    int dp[2*N][M];  //这个数组记得开到2*N,因为遍历后序列长度为2*n-1
    bool vis[N];
    struct edge
    {
        int u,v,w,next;
    }e[2*N];
    
    inline void add(int u ,int v ,int w ,int &k)
    {
        e[k].u = u; e[k].v = v; e[k].w = w; 
        e[k].next = head[u]; head[u] = k++;
        u = u^v; v = u^v; u = u^v;
        e[k].u = u; e[k].v = v; e[k].w = w; 
        e[k].next = head[u]; head[u] = k++;
    }
    
    void dfs(int u ,int dep)
    {
        vis[u] = true; ver[++tot] = u; first[u] = tot; R[tot] = dep;
        for(int k=head[u]; k!=-1; k=e[k].next)
            if( !vis[e[k].v] )
            {
                int v = e[k].v , w = e[k].w;
                dir[v] = dir[u] + w;
                dfs(v,dep+1);
                ver[++tot] = u; R[tot] = dep;
            }
    }
    
    void ST(int len)
    {
        int K = (int)(log((double)len) / log(2.0));
        for(int i=1; i<=len; i++) dp[i][0] = i;
        for(int j=1; j<=K; j++)
            for(int i=1; i+_pow[j]-1<=len; i++)
            {
                int a = dp[i][j-1] , b = dp[i+_pow[j-1]][j-1];
                if(R[a] < R[b]) dp[i][j] = a;
                else            dp[i][j] = b;
            }
    }
    
    int RMQ(int x ,int y)
    {
        int K = (int)(log((double)(y-x+1)) / log(2.0));
        int a = dp[x][K] , b = dp[y-_pow[K]+1][K];
        if(R[a] < R[b]) return a;
        else            return b;
    }
    
    int LCA(int u ,int v)
    {
        int x = first[u] , y = first[v];
        if(x > y) swap(x,y);
        int res = RMQ(x,y);
        return ver[res];
    }
    
    int main()
    {
        for(int i=0; i<M; i++) _pow[i] = (1<<i);
        int cas;
        scanf("%d",&cas);
        while(cas--)
        {
            int n,q,num = 0;
            scanf("%d%d",&n,&q);
            memset(head,-1,sizeof(head));
            memset(vis,false,sizeof(vis));
            for(int i=1; i<n; i++)
            {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                add(u,v,w,num);
            }
            tot = 0; dir[1] = 0;
            dfs(1,1);
            /*
            printf("节点 "); for(int i=1; i<=2*n-1; i++) printf("%d ",ver[i]); cout << endl;
            printf("深度 "); for(int i=1; i<=2*n-1; i++) printf("%d ",R[i]);   cout << endl;
            printf("首位 "); for(int i=0; i<n; i++) printf("%d ",first[i]);    cout << endl;
            printf("距离 "); for(int i=0; i<n; i++) printf("%d ",dir[i]);      cout << endl;
            */
            ST(2*n-1);
            while(q--)
            {
                int u,v;
                scanf("%d%d",&u,&v);
                int lca = LCA(u,v);
    //            printf("lca = %d
    ",lca);
                printf("%d
    ",dir[u] + dir[v] - 2*dir[lca]);
            }
        }
        return 0;
    }

    离线算法:Tarjan算法解决

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    const int N = 40010;
    const int M = 410;
    
    int head[N],__head[N];
    struct edge{
        int u,v,w,next;
    }e[2*N];
    struct ask{
        int u,v,lca,next;
    }ea[M];
    int dir[N],fa[N],ance[N];
    bool vis[N];
    
    inline void add_edge(int u,int v,int w,int &k)
    {
        e[k].u = u; e[k].v = v; e[k].w = w;
        e[k].next = head[u]; head[u] = k++;
        u = u^v; v = u^v; u = u^v;
        e[k].u = u; e[k].v = v; e[k].w = w;
        e[k].next = head[u]; head[u] = k++;
    }
    
    inline void add_ask(int u ,int v ,int &k)
    {
        ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
        ea[k].next = __head[u]; __head[u] = k++;
        u = u^v; v = u^v; u = u^v;
        ea[k].u = u; ea[k].v = v; ea[k].lca = -1;
        ea[k].next = __head[u]; __head[u] = k++;
    }
    
    int Find(int x)
    {
        return x == fa[x] ? x : fa[x] = Find(fa[x]);
    }
    void Union(int u ,int v)
    {
        fa[v] = fa[u];  //可写为  fa[Find(v)] = fa[u];
    }
    
    void Tarjan(int u)
    {
        vis[u] = true;
        ance[u] = fa[u] = u; //课写为 ance[Find(u)] = fa[u] = u;
        for(int k=head[u]; k!=-1; k=e[k].next)
            if( !vis[e[k].v] )
            {
                int v = e[k].v , w = e[k].w;
                dir[v] = dir[u] + w;
                Tarjan(v);
                Union(u,v);
                //ance[Find(u)] = u;  //可写为ance[u] = u;  //甚至不要这个语句都行
            }
        for(int k=__head[u]; k!=-1; k=ea[k].next)
            if( vis[ea[k].v] )
            {
                int v = ea[k].v;
                ea[k].lca = ea[k^1].lca = ance[Find(v)];
            }
    }
    
    int main()
    {
        int cas,n,q,tot;
        scanf("%d",&cas);
        while(cas--)
        {
            scanf("%d%d",&n,&q);
            memset(head,-1,sizeof(head));
            memset(__head,-1,sizeof(__head));
            tot = 0;
            for(int i=1; i<n; i++)
            {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                add_edge(u,v,w,tot);
            }
            tot = 0;
            for(int i=0; i<q; i++)
            {
                int u,v;
                scanf("%d%d",&u,&v);
                add_ask(u,v,tot);
            }
            memset(vis,0,sizeof(vis));
            dir[1] = 0;
            Tarjan(1);
            for(int i=0; i<q; i++)
            {
                int s = i * 2 , u = ea[s].u , v = ea[s].v , lca = ea[s].lca;
                printf("%d
    ",dir[u] + dir[v] - 2*dir[lca]);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4769481.html
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