poj 1986 Distance Queries

LCA

题意：LCA模板题，输入n和m，表示n个点m条边，下面m行是边的信息，两端点和权，后面的那个字母无视掉，没用的。接着k，下面k个询问lca，输出即可

有人说要考虑不连通的情况，我没考虑AC了，另外可能有u，u这样的询问，不过这不影响，照样是写模板，没有特判，一样能过

 

还是Tarjan快一些

 

LCA转RMQ在线算法

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

using namespace std;

#define N 40010

#define M 25



int tot;

int __pow[M];

int head[N];

struct edge{

    int u,v,w,next;

}e[2*N];

int ver[2*N],R[2*N],first[N],dir[N];

int dp[2*N][25];

bool vis[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; first[u] = ++tot; ver[tot] = u; 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 index = RMQ(x,y);

    return ver[index];

}



int main()

{

    for(int i=0; i<M; i++) __pow[i] = (1<<i);

    int n,m,k,str[10];

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        k = 0;

        memset(head,-1,sizeof(head));

        memset(vis,false,sizeof(vis));

        while(m--)

        {

            int u,v,w;

            scanf("%d%d%d%s",&u,&v,&w,str);

            add(u,v,w,k);

        }

        tot = dir[1] = 0;

        dfs(1,1);

        ST(tot);

        int q;

        scanf("%d",&q);

        while(q--)

        {

            int u,v,lca;

            scanf("%d%d",&u,&v);

            lca = LCA(u,v);

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

        }

    }

    return 0;

}

 

Tarjan离线算法

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

#define N 40010

#define M 20010



int head[N];

struct edge{

    int u,v,w,next;

}e[2*N];

int __head[N];

struct ask{

    int u,v,lca,next;

}ea[M];

int fa[N],ance[N],dir[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 Tarjan(int u)

{

    vis[u] = true;

    ance[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);

            fa[v] = u;

        }

    for(int k=__head[u]; k!=-1; k=ea[k].next)

        if( vis[ea[k].v] )

            ea[k].lca = ea[k^1].lca = ance[find(ea[k].v)];

}



int main()

{

    int n,m,q,k; char str[10];

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        memset(head,-1,sizeof(head));

        memset(__head,-1,sizeof(__head));

        memset(vis,false,sizeof(vis));

        k = 0;

        while(m--)

        {

            int u,v,w;

            scanf("%d%d%d%s",&u,&v,&w,str);

            add_edge(u,v,w,k);

        }

        scanf("%d",&q);

        k = 0;

        for(int i=0; i<q; i++)

        {

            int u ,v;

            scanf("%d%d",&u,&v);

            add_ask(u,v,k);

        }

        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;

}