poj 1986 Distance Queries (lca,在线做法dfs+rmq)

题目链接 题意:求树上两点的最短距离? 思路: dis[i]表示点i到根节点的距离,那么任意两点u,v的最短距离d = dis[u]+dis[v]-2*dis[LCA(u,v)]. 只需要求出rmq+dfs的在线方法求出lca(u,v)即可。

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2016年05月20日 星期五 15时36分47秒
  4File Name :code/poj/1986.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=4E4+7;
 34int n,m;
 35vector < pi > edge[N];
 36int q;
 37int in[N];
 38int E[2*N],R[2*N],dis[N],depth[2*N];
 39int p;
 40int dp[2*N][20];
 41void dfs( int u,int dep,int d,int pre)
 42{
 43
 44  //  cout<<"u:"<<u<<" dep:"<<dep<<" d:"<<d<<endl;
 45    p++;
 46    E[p] = u;
 47    depth[p] = dep;
 48    R[u] = p ;
 49    dis[u] = d;
 50
 51
 52    int siz = edge[u].size();
 53    for ( int i = 0 ; i < siz ; i++)
 54    {
 55	int v = edge[u][i].fst;
 56	if (v==pre) continue;
 57	dfs(v,dep+1,d+edge[u][i].sec,u);
 58
 59	p++;
 60	E[p] = u;
 61	depth[p] = dep;
 62    }
 63}
 64
 65
 66
 67int _min( int l,int r)
 68{
 69    if (depth[l]<depth[r]) return l;
 70    return r;
 71}
 72void rmq_init()
 73{
 74    for ( int i = 1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
 75
 76    for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
 77	for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
 78	dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
 79}
 80
 81int rmq_min( int l,int r)
 82{
 83if (l>r) swap(l,r);
 84int k = 0 ;
 85while (1<<(k+1)<=r-l+1) k++;
 86return _min(dp[l][k],dp[r-(1<<k)+1][k]);
 87}
 88int main()
 89{
 90    #ifndef  ONLINE_JUDGE 
 91	freopen("code/in.txt","r",stdin);
 92  #endif
 93
 94	ms(in,0);
 95	scanf("%d %d",&n,&m);
 96	for ( int i = 1 ; i <= n ; i++) edge[i].clear();
 97	for ( int i = 1 ; i <= m ; i++)
 98	{
 99	    int u,v,w;
100	    char dir[5];
101	    scanf("%d%d%d%s",&u,&v,&w,dir);
102	    edge[u].push_back(make_pair(v,w));
103	    edge[v].push_back(make_pair(u,w));
104	}
105
106
107	p = 0 ;
108	dfs(1,0,0,-1);
109	rmq_init();
110
111	scanf("%d",&q);
112	while (q--)
113	{
114	    int u,v;
115	    scanf("%d%d",&u,&v);
116	   // cout<<"u:"<<u<<" v:"<<v<<endl;
117	    int LCA = E[rmq_min(R[u],R[v])];
118	//    cout<<"LCA:"<<LCA<<endl;
119	    int ans;
120	    ans = dis[u]+dis[v]-2*dis[LCA];
121	    printf("%d\n",ans);
122	}
123  #ifndef ONLINE_JUDGE  
124  fclose(stdin);
125  #endif
126    return 0;
127}