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

HintBe 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
[code]#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++)

[/code]



Source
BestCoder Round #43