hdu5266pog loves szh III(RMQ+LCA)

pog loves szh III

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1026    Accepted Submission(s): 232

Problem Description

Pog and Szh are playing games. Firstly Pog draw a tree on the paper. Here we define 1 as the root of the tree.Then Szh choose some nodes from the tree. He wants Pog helps to find the least common ancestor (LCA) of these node.The question is too difficult for Pog.So he decided to simplify the problems.The nodes picked are consecutive numbers from li to ri ([li,ri]) .

Hint : You should be careful about stack overflow !

 

Input

Several groups of data (no more than 3 groups, n≥10000 or Q≥10000 ).

The following line contains ans integers, n(2≤n≤300000) .

AT The following n−1 line, two integers are bi and ci at every line, it shows an edge connecting bi and ci .

The following line contains ans integers, Q(Q≤300000) .

AT The following Q line contains two integers li and ri( 1≤li≤ri≤n ).

 

Output

For each case,output Q integers means the LCA of [li,ri] .

 

Sample Input

   
   
   
   

    
    
    
    5
1 2
1 3
3 4
4 5
5
1 2
2 3
3 4
3 5
1 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
1
3
3
1



    
    
    
    
 
     
 
      Hint 
     
Be careful about stack overflow. 
     
方法一:
 
     
树链剖分维护lca,这样速度比较快,通过RMQ维护连续的LCA值
 
     
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
const int maxn=301010;
int tot;
vector<int>G[maxn]; 
int siz[maxn],son[maxn],fa[maxn],deep[maxn],top[maxn],id[maxn],dp[maxn][20];;

void init(int n){
    tot=0;
    for(int i=1;i<=n;i++)
        G[i].clear();
}

void dfs1(int u,int dep){
    deep[u]=dep;
    siz[u]=1,son[u]=0;
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(v==fa[u])
            continue;
        fa[v]=u;
        dfs1(v,dep+1);
        if(siz[son[u]]<siz[v]){
            son[u]=v;
        }
        siz[u]+=siz[v];
    }
}

void dfs2(int u,int tp){
    top[u]=tp;
    id[u]=++tot;
    if(son[u]!=0){
        dfs2(son[u],tp);
    }
    for(int i=0;i<G[u].size();i++){
        int v=G[u][i];
        if(son[u]==v||v==fa[u])
            continue;
        dfs2(v,v);
    }
}

int Query(int u,int v){
    int f1=top[u],f2=top[v];
    while(f1!=f2){
        if(deep[f1]<deep[f2]){
            swap(f1,f2);
            swap(u,v);
        }
        u=fa[f1];
        f1=top[u];
    }
    if(deep[u]>deep[v])
        return v;
    return u;
}

int RMQ(int L,int R){
    int k=log2(R-L+1);
    return Query(dp[L][k],dp[R-(1<<k)+1][k]);
}

int main(){
    int n,Q;
    while(scanf("%d",&n)!=EOF){
        init(n);
        int u,v;
        for(int i=1;i<n;i++){
            scanf("%d%d",&u,&v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        fa[1]=0,siz[0]=0;
        dfs1(1,0);
        dfs2(1,0);
        for(int i=1;i<=n;i++)
            dp[i][0]=i;
        for(int j=1;(1<<j)<=n;j++)
            for(int i=1;i+(1<<j)-1<=n;i++)
                dp[i][j]=Query(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        scanf("%d",&Q);
        while(Q--){
            scanf("%d%d",&u,&v);
            printf("%d\n",RMQ(u,v));
        }
    }
    return 0;
}
 
     
 
     
 
     
方法二:利用倍增法维护LCA，这样的方法比较慢
 
     
#include<bits/stdc++.h>using namespace std;
const int maxn=300000+1000;
const int DEG=20;
int head[maxn],tot,fa[maxn][DEG];
int dp[maxn][DEG];
int deg[maxn];

struct Edge{
    int to,next;
}e[2*maxn];

void addedge(int u,int v){
    e[tot].to=v;
    e[tot].next=head[u];
    head[u]=tot++;
}

void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}

void bfs(int root){
    fa[root][0]=root;
    deg[root]=0;
    queue<int>Q;
    Q.push(root);
    while(!Q.empty()){
        int tmp=Q.front();
        Q.pop();
        for(int i=1;i<DEG;i++)
            fa[tmp][i]=fa[fa[tmp][i-1]][i-1];//i的第2^j祖先就是i的第2^(j-1)祖先的第2^(j-1)祖先        for(int i=head[tmp];i!=-1;i=e[i].next){
            int v=e[i].to;
            if(fa[tmp][0]==v)   //这个点是tmp的父亲                continue;
            fa[v][0]=tmp;
            deg[v]=deg[tmp]+1;
            Q.push(v);
        }
    }
}

int LCA(int u,int v){
    if(deg[u]>deg[v])
        swap(u,v);
    int hu=deg[u],hv=deg[v];
    int tu=u,tv=v;
    for(int det=hv-hu,i=0;det;det>>=1,i++)
        if(det&1)
            tv=fa[tv][i];
    if(tu==tv)
        return tu;
    for(int i=DEG-1;i>=0;i--){
        if(fa[tu][i]==fa[tv][i])
            continue;
        tu=fa[tu][i];
        tv=fa[tv][i];
    }
    return fa[tu][0];
}

int RMQ(int L,int R){
    int k=log2(R-L+1);
    return LCA(dp[L][k],dp[R-(1<<k)+1][k]);
}

int main(){
    int n,u,v,Q;
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=1;i<n;i++){
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        bfs(1);
        for(int i=1;i<=n;i++)
            dp[i][0]=i;
        for(int j=1;(1<<j)<=n;j++)
 
     
 
     
 
    
    
    
    
     
   
   
   
   

 

Source

BestCoder Round #43