poj 1330 Nearest Common Ancestors (lca,用dfs+rmq在线求解)

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2016年05月19日 星期四 15时05分31秒
  4File Name :code/poj/1330.cpp
  5************************************************ */
  6
  7#include <cstdio>
  8#include <cstring>
  9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define fst first
 20#define sec second
 21#define lson l,m,rt<<1
 22#define rson m+1,r,rt<<1|1
 23#define ms(a,x) memset(a,x,sizeof(a))
 24typedef long long LL;
 25#define pi pair < int ,int >
 26#define MP make_pair
 27
 28using namespace std;
 29const double eps = 1E-8;
 30const int dx4[4]={1,0,0,-1};
 31const int dy4[4]={0,-1,1,0};
 32const int inf = 0x3f3f3f3f;
 33const int N=1E4+7;
 34vector <int>edge[N];
 35int n;
 36int in[N];
 37int cur;
 38int E[2*N];
 39int R[N];
 40int depth[2*N];
 41int dp[2*N][16];
 42void dfs ( int u,int dep)
 43{
 44  //  cout<<"u:"<<u<<" dep:"<<dep<<endl;
 45    cur++;
 46    E[cur] = u;
 47    depth[cur] = dep;
 48    R[u] = cur;  //有向图存的话，在这里访问的一定是第一次经过。
 49
 50    int siz = edge[u].size();
 51    for ( int i = 0 ; i < siz ; i++)
 52    {
 53	int v = edge[u][i];
 54	dfs(v,dep+1);
 55	cur++;  //返回时还经过一次。
 56	E[cur] = u ;
 57	depth[cur] = dep;
 58    }
 59}
 60
 61int _min(int l,int r)
 62{
 63    if (depth[l]<depth[r]) return l;
 64    return r;
 65}
 66
 67void rmq_init()
 68{
 69    for ( int i =  1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
 70
 71    for ( int j = 1 ;  (1<<j) <= 2*n+2 ; j++ )
 72	for ( int i = 1; i + (1<<j)-1 <=2*n+2 ; i++)
 73	    dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
 74}
 75
 76int rmq_min( int l,int r)
 77{
 78    if (l>r) swap(l,r);
 79    int k = 0 ;
 80    while (1<<(k+1)<=r-l+1) k++;
 81    return _min(dp[l][k],dp[r-(1<<k)+1][k]);
 82}
 83int main()
 84{
 85	#ifndef  ONLINE_JUDGE 
 86	freopen("code/in.txt","r",stdin);
 87  #endif
 88	int T;
 89	cin>>T;
 90	while (T--)
 91	{
 92	    scanf("%d",&n);
 93	    ms(in,0);
 94	    ms(E,0);
 95	    ms(R,0);
 96	    ms(depth,0);
 97	    for ( int i = 1 ; i <= n ; i++) edge[i].clear();
 98	    for ( int i = 1 ; i <= n-1 ; i++)
 99	    {
100		int u,v;
101		scanf("%d %d",&u,&v);
102		edge[u].push_back(v);
103		in[v]++;
104	    }
105
106	    int root;
107	    for ( int i = 1 ; i <= n ; i++) if (in[i]==0) root = i;
108
109	    cur = 0 ;
110	    dfs(root,0);
111	    rmq_init();
112
113	    int x,y;
114	    scanf("%d %d",&x,&y);
115	    printf("%d\n",E[rmq_min(R[x],R[y])]);
116	}
117
118  #ifndef ONLINE_JUDGE  
119  fclose(stdin);
120  #endif
121    return 0;
122}