pku Nearest Common Ancestors LCA 问题（rmq ＆＆ tarjan）解决

http://poj.org/problem?id=1330

题意：给n个点，n-1条边，一棵树，求每个询问的LCA;

思路：

rmq求LCA。。。。http://dongxicheng.org/structure/lca-rmq/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 10007
using namespace std;
//freopen("din.txt","r",stdin);

struct node{
    int v;
    int next;
}g[N];
int head[N],ct;

int E[2*N],R[N],D[2*N];
int dp[2*N][20],f[N],pow2[21];
bool vt[N];
int n,cnt;

void init(){
    for (int i = 0; i <= n; ++i){
        f[i] = -1;
        head[i] = -1;
    }
    ct = 0; cnt = 0;
    CL(vt,false);
}
void add(int u,int v){
    g[ct].v = v;
    g[ct].next = head[u];
    head[u] = ct++;
}
//E,R,D对应的值
void dfs(int u,int dep){
    vt[u] = true;
    E[cnt] = u;
    R[u] = cnt;
    D[cnt++] = dep;
    for (int i = head[u]; i != -1; i = g[i].next){
        int v = g[i].v;
        if (!vt[v]){
            dfs(v,dep + 1);
            E[cnt] = u;
            D[cnt++] = dep;
        }
    }
}
int MIN(int i,int j){
    if (D[i] < D[j]) return i;
    else return j;
}
void init_rmq(){

    int nn = 2*n - 1;
    int i,j;
    for (i = 0; i < nn; ++i)
        dp[i][0] = i;

    for (j = 1; pow2[j] <= nn; ++j){
        for (i = 0; (i + pow2[j] - 1) < nn; ++i)
        dp[i][j] = MIN(dp[i][j - 1],dp[i + pow2[j - 1]][j  - 1]);
    }
}
int rmq(int l,int r){
    int k = (int)(log(1.0*(r - l + 1))/log(2.0));
    return MIN(dp[l][k],dp[r - pow2[k] + 1][k]);
}
int main(){
    //freopen("din.txt","r",stdin);
    int T,i;
    int x,y;
    for (i = 0; i < 20; ++i) pow2[i] = 1<<i;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        init();
        for (i = 1; i < n; ++i){
            scanf("%d%d",&x,&y);
            add(x,y);
            f[y] = x;
        }
        //找出根节点
        for (i = 1; i <= n; ++i){
            if (f[i] == -1){
                break;
            }
        }
        cnt = 0;
        dfs(i,0);
        init_rmq();

        scanf("%d%d",&x,&y);
        if (R[x] <= R[y]) printf("%d\n",E[rmq(R[x],R[y])]);
        else printf("%d\n",E[rmq(R[y],R[x])]);
    }
    return 0;
}

 

 tarjan离线处理：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <string>

#define CL(a,num) memset((a),(num),sizeof(a))
#define iabs(x)  ((x) > 0 ? (x) : -(x))
#define Min(a,b) (a) > (b)? (b):(a)
#define Max(a,b) (a) > (b)? (a):(b)

#define ll __int64
#define inf 0x7f7f7f7f
#define MOD 100000007
#define lc l,m,rt<<1
#define rc m + 1,r,rt<<1|1
#define pi acos(-1.0)
#define test puts("<------------------->")
#define maxn 100007
#define M 100007
#define N 10007
using namespace std;
//freopen("din.txt","r",stdin);

struct node{
    int v;
    int next;
}g[N];

int head[N],ct;
bool vt[N];
vector<int>ask[N];
int f[N],pa[N],pre[N];
int n;

int find(int x){
    if (x != f[x]) f[x] = find(f[x]);
    return f[x];
}
void Union(int x,int y){
    x = find(x);
    y = find(y);
    f[x] = y;
}
void add(int u,int v){
    g[ct].v = v;
    g[ct].next = head[u];
    head[u] = ct++;
}
void LCA(int u){
    int i;
    f[u] = u;//建立以当前节点为根的树
    for (i = head[u]; i != -1; i = g[i].next){
        int v = g[i].v;
        LCA(v);
        Union(u,v);//记住是u指向v
        pa[find(u)] = u;//记录该树的父节点
    }

    vt[u] = true;

    int sz = ask[u].size();
    for (i = 0; i < sz; ++i){
        if (vt[ask[u][i]]){//如果同时访问就可求出lca
            printf("%d\n",pa[find(ask[u][i])]);
        }
    }
}
int main(){
    //freopen("din.txt","r",stdin);
    int i;
    int T,x,y;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        ct = 0; CL(head,-1); CL(pre,-1);
        for (i = 1; i < n; ++i){
            scanf("%d%d",&x,&y);
            add(x,y);
            pre[y] = x;//记录父节点主要在于找根
        }

        for (i = 1; i <= n; ++i){
            f[i] = i;
            ask[i].clear();//离线处理
        }

        scanf("%d%d",&x,&y);
        ask[x].push_back(y);
        ask[y].push_back(x);

        CL(vt,false);
        //找父节点
        for (i = 1; i <= n; ++i){
            if (pre[i] == -1) break;
        }
        LCA(i);
    }
    return 0;
}