POJ 1470 Closest Common Ancestors（离线tarjan-LCA）

Description
给出一棵节点数为n的树和q次查询，每次查询a和b的LCA，最后输出每个节点被查询的次数
Input
第一行为一整数n表示树的节点数，之后n行每行输入一个节点的邻接关系，然后是一整数q表示查询次数，最后q行每行两个整数a和b表示查询a和b的LCA
Output
如果某个节点作为查询中两个点的LCA，则输出其被查询的次数
Sample Input
5
5:(3) 1 4 2
1:(0)
4:(0)
2:(1) 3
3:(0)
6
(1 5) (1 4) (4 2)
(2 3)
(1 3) (4 3)
Sample Output
2:1
5:5
Solution
离线所有查询并求出LCA后统计次数即可
Code

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define maxn 1111#define maxq 555555struct Edge{    int to,next;}edge[maxn*2];struct Query{    int to,next,id;}query[maxq*2];int f[maxn];bool vis[maxn];int ancestor[maxn];int head1[maxn],tot1;int head2[maxq],tot2;int lca[maxq]; void init(){    tot1=tot2=0;    memset(head1,-1,sizeof(head1));    memset(head2,-1,sizeof(head2));    memset(vis,0,sizeof(vis));    memset(f,-1,sizeof(f));    memset(ancestor,0,sizeof(ancestor));}int find(int x){    if(f[x]==-1)return x;    return f[x]=find(f[x]);}void unite(int x,int y){    x=find(x),y=find(y);    if(x!=y)f[x]=y;}void add_edge(int u,int v){    edge[tot1].to=v;    edge[tot1].next=head1[u];    head1[u]=tot1++;}void add_query(int u,int v,int id){    query[tot2].to=v;    query[tot2].next=head2[u];    query[tot2].id=id;    head2[u]=tot2++;    query[tot2].to=u;    query[tot2].next=head2[v];    query[tot2].id=id;    head2[v]=tot2++;}void tarjan(int u){    ancestor[u]=u;    vis[u]=1;    for(int i=head1[u];~i;i=edge[i].next)    {        int v=edge[i].to;        if(vis[v])continue;        tarjan(v);        unite(u,v);        ancestor[find(u)]=u;    }    for(int i=head2[u];~i;i=query[i].next)    {        int v=query[i].to;        if(vis[v])lca[query[i].id]=ancestor[find(v)];    }} int main(){    int n,q,root,flag[maxn],ans[maxn];    while(~scanf("%d",&n))    {        init();        memset(flag,0,sizeof(flag));        memset(ans,0,sizeof(ans));        int u,v,num;        for(int i=1;i<=n;i++)        {            scanf("%d:(%d)",&u,&num);            while(num--)            {                scanf("%d",&v);                add_edge(u,v),add_edge(v,u);                flag[v]=1;            }        }        scanf("%d",&q);        for(int i=1;i<=q;i++)        {            char ch;            cin>>ch;            scanf("%d %d)",&u,&v);            add_query(u,v,i);        }        for(int i=1;i<=n;i++)            if(!flag[i])            {                root=i;                break;            }        tarjan(root);        for(int i=1;i<=q;i++)            ans[lca[i]]++;        for(int i=1;i<=n;i++)            if(ans[i])printf("%d:%d\n",i,ans[i]);    }    return 0;}