【离线Tarjan 模板】LCA tarjan 算法 练习： hdu 2586 + poj 1986

个人一直觉得kuangbin大牛的LCA板子 感觉有点不太好用，于是在这个大牛的博客帮助下，更换了自己的板子；
http://www.cnblogs.com/scau20110726/archive/2013/05/26/3100265.html

hdu 2586，
给一颗树 ，每一条边有权值，然后m次询问，每一次询问两个点：i，j。 输出 i，j的最短距离。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N = 40010;const int M = 410;int head[N],__head[N]; //同理 一个链式存 树，另外一个链式存问题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)];        }}void init(){      memset(head,-1,sizeof(head));      memset(__head,-1,sizeof(__head));      memset(vis,0,sizeof(vis));}int main(){    int cas,n,q,tot;    scanf("%d",&cas);    while(cas--)    {        init();        scanf("%d%d",&n,&q);        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);        }        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;}

poj 1986
题意：给一棵树，询问m次，也是输出他们的最短距离，这题貌似他们判断什么连通不连通，询问点相同不相同，但是用了这个模板啥都不用判断，直接过。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int N = 40010;const int M = 40010;int head[N],__head[N]; //同理 一个链式存 树，另外一个链式存问题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)];        }}void init(){      memset(head,-1,sizeof(head));      memset(__head,-1,sizeof(__head));      memset(vis,0,sizeof(vis));}int main(){    int n,m;    char ch[5];    while(~scanf("%d %d",&n,&m)){        init();        int tot=0;        for(int i=0;i<m;i++){            int u,v,w;            scanf("%d %d %d %s",&u,&v,&w,ch);            add_edge(u,v,w,tot);        }        tot=0;        int q;        scanf("%d",&q);        for(int i=0;i<q;i++){            int u,v;            scanf("%d %d",&u,&v);            add_ask(u,v,tot);        }        dir[1]=0;        Tarjan(1);        for(int i=0;i<q;i++){            int s=2*i;            int a=ea[s].u,b=ea[s].v,c=ea[s].lca;//          printf("%d %d %d\n",a,b,c);            int ans=dir[a]+dir[b]-2*dir[c];            printf("%d\n",ans);        }    }    return 0;}