poj 1470：Closest Common Ancestors（最近公共祖先，Tarjan模版）

大致题意：
给出一棵根树，给出一些成对的点，求出每对点的最近公共祖先。
大致思路：
第一道Tarjan，很裸，这个题目算是当模版用了，代码修改自 子任博客

详细代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define MAXN 1100
#define MAXM 500000
struct Edge{
    int u,v,next;
} e[2*MAXN];
struct Query{
    int u,v,w,next;
} e1[MAXM];
int n, m, cnt1, cnt2;
int head1[MAXN], head2[MAXN], parent[MAXN], ans[MAXM], ret[MAXN];
bool vis[MAXN];
bool use[MAXN];
void init(){
    cnt1 = cnt2 = 1;
    memset(head1,-1,sizeof(int)*(n+1));
    memset(head2,-1,sizeof(int)*(n+1));
    memset(vis,false,sizeof(bool)*(n+1));
    memset(parent,-1,sizeof(int)*(n+1));
}
void addedge(int u, int v){
    e[cnt1].u = u;
    e[cnt1].v = v;
    e[cnt1].next = head1[u];
    head1[u] = cnt1++;
}
void addedge2(int u,int v,int w){
    e1[cnt2].u = u;
    e1[cnt2].v = v;
    e1[cnt2].w = w;
    e1[cnt2].next = head2[u];
    head2[u] = cnt2++;
}
int find(int u){
    if(parent[u] == -1)
        return u;
    return parent[u] = find(parent[u]);
}
void tarjan(int u){
    int i, v, w;
    vis[u] = true;
    for(i = head2[u]; i != -1; i = e1[i].next){
        v = e1[i].v;
        w = e1[i].w;
        if(vis[v])
            ans[w] = find(v);
    }
    for(i = head1[u]; i != -1; i = e[i].next){
        v = e[i].v;
        if(!vis[v]){
            tarjan(v);
            parent[v] = u;
        }
    }
}
int main(){
    int i, j, u, v;
    while(scanf("%d",&n) != -1){
        init();
        memset(use,false,sizeof(bool)*(n+1));
        for(i = 1; i <= n; i++){
            scanf("%d:",&u);
            while(getchar() != '(');
            scanf("%d",&j);
            while(getchar() != ')');
            while(j--){
                scanf("%d",&v);
                addedge(u,v);
                use[v] = true;
            }
        }
        scanf("%d",&m);
        for(i = 1; i <= m; i++){
            while(getchar() != '(');
            scanf("%d%d",&u,&v);
            addedge2(u,v,i);
            addedge2(v,u,i);
            while(getchar() != ')');
        }
        for(i = 1; i <= n; i++)  //寻找根节点，这里是trick
            if(!use[i])
                break;
        tarjan(i);
        sort(ans+1,ans+m+1);     ////ans[i]用来存放第i问中两点的公共祖先
        memset(ret,0,sizeof(ret));
        for(i = 1; i <= m; i++){
            ret[ans[i]]++;
        }
        for(i =1; i <= n; i++)
            if(ret[i]!= 0)printf("%d:%d\n",i,ret[i]);
    }
    return 0;
}