poj 1986 Distance Queries

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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

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#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

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#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;
}