hdu 2586 How far away ？

LCA模板题

 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\n",lca);

            printf("%d\n",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\n",dir[u] + dir[v] - 2*dir[lca]);

        }

    }

    return 0;

}