POJ 1470 Closest Common Ancestors (LCA, dfs+ST在线算法)
| Time Limit: 2000MS | Memory Limit: 10000K | |
| Total Submissions: 13370 | Accepted: 4338 |
Description
Input
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 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
Source
模板题
/* ***********************************************
Author :kuangbin
Created Time :2013-9-5 8:54:16
File Name :F:\2013ACM练习\专题学习\LCA\POJ1470.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = ;
int rmq[*MAXN];//rmq数组,就是欧拉序列对应的深度序列
struct ST
{
int mm[*MAXN];
int dp[*MAXN][];//最小值对应的下标
void init(int n)
{
mm[] = -;
for(int i = ;i <= n;i++)
{
mm[i] = ((i&(i-)) == )?mm[i-]+:mm[i-];
dp[i][] = i;
}
for(int j = ; j <= mm[n];j++)
for(int i = ; i + (<<j) - <= n; i++)
dp[i][j] = rmq[dp[i][j-]] < rmq[dp[i+(<<(j-))][j-]]?dp[i][j-]:dp[i+(<<(j-))][j-];
}
int query(int a,int b)//查询[a,b]之间最小值的下标
{
if(a > b)swap(a,b);
int k = mm[b-a+];
return rmq[dp[a][k]] <= rmq[dp[b-(<<k)+][k]]?dp[a][k]:dp[b-(<<k)+][k];
}
};
//边的结构体定义
struct Edge
{
int to,next;
};
Edge edge[MAXN*];
int tot,head[MAXN]; int F[MAXN*];//欧拉序列,就是dfs遍历的顺序,长度为2*n-1,下标从1开始
int P[MAXN];//P[i]表示点i在F中第一次出现的位置
int cnt; ST st;
void init()
{
tot = ;
memset(head,-,sizeof(head));
}
void addedge(int u,int v)//加边,无向边需要加两次
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int u,int pre,int dep)
{
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for(int i = head[u];i != -;i = edge[i].next)
{
int v = edge[i].to;
if(v == pre)continue;
dfs(v,u,dep+);
F[++cnt] = u;
rmq[cnt] = dep;
}
}
void LCA_init(int root,int node_num)//查询LCA前的初始化
{
cnt = ;
dfs(root,root,);
st.init(*node_num-);
}
int query_lca(int u,int v)//查询u,v的lca编号
{
return F[st.query(P[u],P[v])];
}
bool flag[MAXN];
int Count_num[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int u,v,k;
int Q;
while(scanf("%d",&n) == )
{
init();
memset(flag,false,sizeof(flag));
for(int i = ;i <= n;i++)
{
scanf("%d:(%d)",&u,&k);
while(k--)
{
scanf("%d",&v);
flag[v] = true;
addedge(u,v);
addedge(v,u);
}
}
int root;
for(int i = ;i <= n;i++)
if(!flag[i])
{
root = i;
break;
}
LCA_init(root,n);
memset(Count_num,,sizeof(Count_num));
scanf("%d",&Q);
while(Q--)
{
char ch;
cin>>ch;
scanf("%d %d)",&u,&v);
Count_num[query_lca(u,v)]++;
}
for(int i = ;i <= n;i++)
if(Count_num[i] > )
printf("%d:%d\n",i,Count_num[i]);
}
return ;
}
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