Closest Common Ancestors

Description

Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)

Input

The data set, which is read from a the std input, starts with the tree description, in the form: 

nr_of_vertices 

vertex:(nr_of_successors) successor1 successor2 … successorn 

…

where vertices are represented as integers from 1 to n ( n <= 900 ). The tree description is followed by a list of pairs of vertices, in the form: 

nr_of_pairs 

(u v) (x y) … 

The input file contents several data sets (at least one). 

Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

Output

For each common ancestor the program prints the ancestor and the number of pair for which it is an ancestor. The results are printed on the standard output on separate lines, in to the ascending order of the vertices, in the format: ancestor:times 

For example, for the following tree: 

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

Hint

Huge input, scanf is recommended.

Source

Southeastern Europe 2000

模板题

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013-9-5 8:54:16
  4 File Name     :F:\2013ACM练习\专题学习\LCA\POJ1470.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN = 1010;
 21 int rmq[2*MAXN];//rmq数组，就是欧拉序列对应的深度序列
 22 struct ST
 23 {
 24     int mm[2*MAXN];
 25     int dp[2*MAXN][20];//最小值对应的下标
 26     void init(int n)
 27     {
 28         mm[0] = -1;
 29         for(int i = 1;i <= n;i++)
 30         {
 31             mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
 32             dp[i][0] = i;
 33         }
 34         for(int j = 1; j <= mm[n];j++)
 35             for(int i = 1; i + (1<<j) - 1 <= n; i++)
 36                 dp[i][j] = rmq[dp[i][j-1]] < rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
 37     }
 38     int query(int a,int b)//查询[a,b]之间最小值的下标
 39     {
 40         if(a > b)swap(a,b);
 41         int k = mm[b-a+1];
 42         return rmq[dp[a][k]] <= rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];
 43     }
 44 };
 45 //边的结构体定义
 46 struct Edge
 47 {
 48     int to,next;
 49 };
 50 Edge edge[MAXN*2];
 51 int tot,head[MAXN];
 52 
 53 int F[MAXN*2];//欧拉序列，就是dfs遍历的顺序，长度为2*n-1,下标从1开始
 54 int P[MAXN];//P[i]表示点i在F中第一次出现的位置
 55 int cnt;
 56 
 57 ST st;
 58 void init()
 59 {
 60     tot = 0;
 61     memset(head,-1,sizeof(head));
 62 }
 63 void addedge(int u,int v)//加边，无向边需要加两次
 64 {
 65     edge[tot].to = v;
 66     edge[tot].next = head[u];
 67     head[u] = tot++;
 68 }
 69 void dfs(int u,int pre,int dep)
 70 {
 71     F[++cnt] = u;
 72     rmq[cnt] = dep;
 73     P[u] = cnt;
 74     for(int i = head[u];i != -1;i = edge[i].next)
 75     {
 76         int v = edge[i].to;
 77         if(v == pre)continue;
 78         dfs(v,u,dep+1);
 79         F[++cnt] = u;
 80         rmq[cnt] = dep;
 81     }
 82 }
 83 void LCA_init(int root,int node_num)//查询LCA前的初始化
 84 {
 85     cnt = 0;
 86     dfs(root,root,0);
 87     st.init(2*node_num-1);
 88 }
 89 int query_lca(int u,int v)//查询u,v的lca编号
 90 {
 91     return F[st.query(P[u],P[v])];
 92 }
 93 bool flag[MAXN];
 94 int Count_num[MAXN];
 95 int main()
 96 {
 97     //freopen("in.txt","r",stdin);
 98     //freopen("out.txt","w",stdout);
 99     int n;
100     int u,v,k;
101     int Q;
102     while(scanf("%d",&n) == 1)
103     {
104         init();
105         memset(flag,false,sizeof(flag));
106         for(int i = 1;i <= n;i++)
107         {
108             scanf("%d:(%d)",&u,&k);
109             while(k--)
110             {
111                 scanf("%d",&v);
112                 flag[v] = true;
113                 addedge(u,v);
114                 addedge(v,u);
115             }
116         }
117         int root;
118         for(int i = 1;i <= n;i++)
119             if(!flag[i])
120             {
121                 root = i;
122                 break;
123             }
124         LCA_init(root,n);
125         memset(Count_num,0,sizeof(Count_num));
126         scanf("%d",&Q);
127         while(Q--)
128         {
129             char ch;
130             cin>>ch;
131             scanf("%d %d)",&u,&v);
132             Count_num[query_lca(u,v)]++;
133         }
134         for(int i = 1;i <= n;i++)
135             if(Count_num[i] > 0)
136                 printf("%d:%d\n",i,Count_num[i]);
137     }
138     return 0;
139 }