POJ 1470 Closest Common Ancestors (模板题)(Tarjan离线)【LCA】

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<vector>
 5 
 6 using namespace std;
 7 const int N = 1100;
 8 
 9 vector<int>edge[N];
10 int query[N][N],father[N],count[N],indeg[N];
11 bool vis[N];
12 int n,m;
13 void init(){
14     for(int i=1;i<=n;i++)edge[i].clear();
15     memset(query,0,sizeof(query));
16     memset(vis,false,sizeof(vis));
17     memset(count,0,sizeof(count));
18     memset(indeg,0,sizeof(indeg));
19 }
20 int find(int x){    //找到根节点
21     if(x!=father[x])
22         father[x]=find(father[x]);
23     return father[x];
24 }
25 void Tarjan(int u){
26     father[u]=u;
27     for(int i=0;i<edge[u].size();i++){   //得到该树上所有节点的父子关系
28         int v=edge[u][i];
29         Tarjan(v);
30         father[v]=u;
31     }
32     vis[u]=true;
33     for(int i=1;i<=n;i++)
34         if(vis[i] && query[u][i])
35             count[find(i)]+=query[u][i];   //最近公共祖先出现次数+1
36 }
37 int main(){
38     while(~scanf("%d",&n)){
39         init();
40         int u,v;
41         for(int i=0;i<n;i++){
42             scanf("%d:(%d)",&u,&m);
43             while(m--){
44                 scanf(" %d",&v);
45                 edge[u].push_back(v);   //建立有向边
46                 indeg[v]++;    //统计入度，用于寻找根节点
47             }
48         }
49         scanf(" %d",&m);
50         for(int i=0;i<m;i++){
51             scanf(" (%d %d)",&u,&v);
52             query[u][v]++;     //将代查询的节点也全部记录下来
53             query[v][u]++;
54         }
55         for(int i=1;i<=n;i++)
56             if(indeg[i]==0){
57                 Tarjan(i);
58                 break;
59             }
60         for(int i=1;i<=n;i++)
61             if(count[i])
62                 printf("%d:%d\n",i,count[i]);
63     }
64     return 0;
65 }