Online LCA - Using RMQ


LCA与RMQ互相转化 » NoAlGo博客
http://www.zhihu.com/question/19957473
2. O(n log n) 预处理 + O(1) 查询的在线LCA算法
这个算法不要求事先知道所有查询。来一个查询就处理一个,每处理一个复杂度是常数。 其思想是将LCA问题转化成RMQ(Range Minimum/Maximum Query),然后再利用高效的RMQ算法如线段树 ,Sparse Table 等求解。对于RMQ问题,如果要追求O(1)的查询复杂度,就不能用线段树了(线段树是O(log n)的),所以可以采用基于动态规划的Sparse Table方法。

(1)将LCA 转化成RMQ:
对于有n个节点的树,我们先建三个一维数组: First[0...n], Depth[0...2*n], Seq[0, 2*n](稍后解释这三个数组的意义)。然后DFS这棵树,传统的打印DFS序列的流程如下:
function DFS(u)
     u.visit = true
     print u.id
     foreach child v of u
          if v.visit == false
              dfs(v)
          end
     end
end

我们将上述流程修改一下:
function DFS(u)
     u.visit = true
     print u.id
     foreach child v of u
         if v.visit == false
             dfs(v)
         end
         print u.id
     end
end

也就是在每次回溯时我们也把节点u打印一下。可以证明这时的dfs序列长度是2*n-1(传统dfs序列的长度是n

LCA转RMQ主要是通过DFS(深度优先搜索)完成,时间复杂度为O(n)。
DFS时每次进入以及回溯到某一个节点时,我们都记录访问到的节点val,同时记录下这个节点的深度depth。因为进入及回溯时都要记录,所以一个节点可能会被记录多次。另外,为了后面应用RMQ的方便,还使用数组first记录每个元素第一次访问的下标。
http://blog.csdn.net/smallacmer/article/details/7432664
LCA与RMQ问题的转化,LCA问题可以经过DFS+st变成rmq(min)的解法,而且求解的时间复杂度规模是一样的.

DFS时每次进入以及回溯到某一个节点时,我们都记录访问到的节点val,同时记录下这个节点的深度depth。因为进入及回溯时都要记录,所以一个节点可能会被记录多次。另外,为了后面应用RMQ的方便,还使用数组first记录每个元素第一次访问的下标。
例如,下面是一棵以节点0为根节点的多叉树:
lca例子树
以这棵树为例,我们进行DFS时可以得到如下depth数组和val数组:
val014157510203630
depth012123210101210
first数组为
下标01234567
0191124125
要求x和y的LCA时,因为LCA的深度肯定是最小的,于是可以通过first数组找到x和y首次出现的位置first[x]和first[y],然后再depth数组查找从first[x]到first[y]的RMQ的下标id,然后在val数组中找到下标id所代表的元素,即为x和y的LCA。
即 LCA(x, y) = val[RMQ(depth, first[x], first[y])]
int depth[mx], val[mx], first[mx];
int id; //记录当前数组的长度
void dfs(int x, int dep)
{
 depth[id] = dep; val[id] = x; first[x] = id; id++;
 for (int i = 0; i < tree[x].size(); i++)
 {
  dfs(tree[x][i], dep+1);
  depth[id] = dep; val[id] = x; id++; //递归完一棵子树回到自身
 }
}
https://gist.github.com/ddrone/2975702
Implementation of online static RMQ-problem with O(N) preprocessing time and O(1) query time Raw

http://kmplayer.iteye.com/blog/604232
  1. struct node //可以添加额外的信息  
  2. {  
  3.     int v;//孩子结点  
  4. };  
  5.   
  6. //注意vector在树问题中的使用  
  7. vector<node> tree[maxn];  
  8.   
  9. int dfsnum[maxn]; //记录遍历的节点  
  10. int depth[maxn]; //记录节点对应的深度  
  11. int first[maxn]; //记录结点第一次访问到时的下标  
  12. int top; //记录总的步伐数  
  13.   
  14. void dfs(int m,int f,int dep) //当前节点编号,父节点编号,深度  
  15. {  
  16.     dfsnum[top]=m;  
  17.     depth[top]=dep;  
  18.     first[m]=top;  
  19.     top++;  
  20.     for(unsigned i=0;i<tree[m].size();i++)  
  21.     {  
  22.         if(tree[m][i].v==f)  
  23.             continue;  
  24.         dfs(tree[m][i].v,m,dep+1);  
  25.         dfsnum[top]=m; //注:每条边回溯一次,所以top的值=n+n-1  
  26.         depth[top]=dep;  
  27.         top++;  
  28.     }  
  29. }  
  30.   
  31. int dp[maxn][18];  
  32. void makeRmqIndex(int n,int b[]) //返回最小值对应的下标  
  33. {  
  34.     int i,j;  
  35.     for(i=0;i<n;i++)  
  36.         dp[i][0]=i;  
  37.     for(j=1;(1<<j)<=n;j++)  
  38.         for(i=0;i+(1<<j)-1<n;i++)  
  39.             dp[i][j]=b[dp[i][j-1]] < b[dp[i+(1<<(j-1))][j-1]]? dp[i][j-1]:dp[i+(1<<(j-1))][j-1];  
  40. }  
  41. int rmqIndex(int s,int v,int b[])  
  42. {  
  43.     int k=(int)(log((v-s+1)*1.0)/log(2.0));  
  44.     return b[dp[s][k]]<b[dp[v-(1<<k)+1][k]]? dp[s][k]:dp[v-(1<<k)+1][k];  
  45. }  
  46.   
  47. int lca(int x,int y)  
  48. {  
  49.     return dfsnum[rmqIndex(first[x],first[y],depth)];  
  50. }  
  51.   
  52. int main()  
  53. {  
  54.     int n=5;//顶点数  
  55.     top=0;  
  56.     //分别存放每条边的端点  
  57.     int x[]={1,1,3,3};  
  58.     int y[]={2,3,4,5};  
  59.     node temp;  
  60.     for(int i=0;i<n-1;i++) //n-1条边  
  61.     {  
  62.         temp.v=y[i];  
  63.         tree[x[i]].push_back(temp);  
  64.         temp.v=x[i];  
  65.         tree[y[i]].push_back(temp);  
  66.     }  
  67.     dfs(1,-1,0); //根节点为1  
  68.     cout<<"总数:"<<top<<endl;  
  69.     makeRmqIndex(top,depth);   
  70.     cout<<"lca(4,5):"<<lca(4,5)<<endl; 
http://www.codeproject.com/Articles/141999/Solving-LCA-by-Reducing-to-RMQ-in-C
Read full article from LCA与RMQ互相转化 » NoAlGo博客

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