最近公共祖先LCA【模板】

Tarjan-LCA算法：
对于每一点u：
1.建立以u为代表元素的集合。
2.遍历与u相连的节点v，如果没有被访问过，对于v使用Tarjan-LCA算法，结束后，将v的集合并入u的集合。
3.对于与节点u相关的询问(u，v)，如果v被访问过，则结果就是v所在集合的所代表的元素。

求(u,v)的最近公共祖先节点，则询问时调用QEdges[k].lca = find(QEdges[k].to);
求(u,v)在树上的距离，Dist(u，v) = Dist(1，u) + Dist(1，v) - 2*Dist( 1，LCA(u，v) )，即u到根结点的距离 + v到根结点的距离 - 2*(u，v)最近公共祖先到根结点的距离，则询问时调用QEdges[k].lca = Dist[u] + Dist[QEdges[k].to] - 2*Dist[find(QEdges[k].to)];

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int MAXN = 80080;const int MAXQ = 20020;int father[MAXN],Head[MAXN],QHead[MAXN],Dist[MAXN];//Head[]和Eges[]用来存储原图；QHead[]和QEdges[]用来存储询问struct EdgeNode{    int to;    int next;    int lca;}Edges[MAXN],QEdges[MAXN];int find(int x){    if(x != father[x])        father[x] = find(father[x]);    return father[x];}bool vis[MAXN];void LCA(int u){    father[u] = u;    vis[u] = true;    for(int k = Head[u]; k != -1; k = Edges[k].next)    {        if(!vis[Edges[k].to])        {            Dist[Edges[k].to] = Dist[u] + Edges[k].lca;            LCA(Edges[k].to);            father[Edges[k].to] = u;        }    }    for(int k = QHead[u]; k != -1; k = QEdges[k].next)    {        if(vis[QEdges[k].to])        {            //QEdges[k].lca = find(QEdges[k].to);            QEdges[k].lca = Dist[u] + Dist[QEdges[k].to] - 2*Dist[find(QEdges[k].to)];            QEdges[k^1].lca = QEdges[k].lca;        }    }}int main(){    int N,M,K,u,v,w,a,b;    while(~scanf("%d%d",&N,&M))    {        memset(father,0,sizeof(father));        memset(Head,-1,sizeof(Head));        memset(QHead,-1,sizeof(QHead));        memset(vis,false,sizeof(vis));        memset(Edges,0,sizeof(Edges));        memset(QEdges,0,sizeof(QEdges));        memset(Dist,0,sizeof(Dist));        int id = 0;        for(int i = 0; i < M; ++i)//插入图的M条边        {               scanf("%d%d%d",&u,&v,&w);            Edges[id].to = v;            Edges[id].lca = w;            Edges[id].next = Head[u];            Head[u] = id++;            Edges[id].to = u;            Edges[id].lca = w;            Edges[id].next = Head[v];            Head[v] = id++;        }   //(u,v)和(v,u)都要存，表示双向边        scanf("%d",&K);//K条询问        int iq = 0;        for(int i = 0; i < K; ++i)        {            scanf("%d%d",&a,&b);            QEdges[iq].to = b;            QEdges[iq].next = QHead[a];            QHead[a] = iq++;            QEdges[iq].to = a;            QEdges[iq].next = QHead[b];            QHead[b] = iq++;        }   //同理(u,v)和(v,u)都要存，但是询问时只对一条边回答        LCA(1); //跟结点。        for(int i = 0; i < iq; i+=2)    //回答询问            printf("%d\n",QEdges[i].lca);    }    return 0;}