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。

/* ***********************************************
Author :111qqz
Created Time :2016年05月20日 星期五 15时36分47秒
File Name :code/poj/1986.cpp
************************************************ */
 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=4E4+7;
 7int n,m;
 8vector < pi > edge[N];
 9int q;
10int in[N];
11int E[2*N],R[2*N],dis[N],depth[2*N];
12int p;
13int dp[2*N][20];
14void dfs( int u,int dep,int d,int pre)
15{
1  //  cout<<"u:"<<u<<" dep:"<<dep<<" d:"<<d<<endl;
2    p++;
3    E[p] = u;
4    depth[p] = dep;
5    R[u] = p ;
6    dis[u] = d;
1    int siz = edge[u].size();
2    for ( int i = 0 ; i < siz ; i++)
3    {
4	int v = edge[u][i].fst;
5	if (v==pre) continue;
6	dfs(v,dep+1,d+edge[u][i].sec,u);
1	p++;
2	E[p] = u;
3	depth[p] = dep;
4    }
5}
1int _min( int l,int r)
2{
3    if (depth[l]<depth[r]) return l;
4    return r;
5}
6void rmq_init()
7{
8    for ( int i = 1 ; i <= 2*n+2 ; i++) dp[i][0] = i;
1    for ( int j = 1 ; (1<<j) <= 2*n+2 ; j++)
2	for ( int i = 1 ; i + (1<<j)-1 <= 2*n+2 ; i++)
3	dp[i][j] = _min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
4}
 1int rmq_min( int l,int r)
 2{
 3if (l>r) swap(l,r);
 4int k = 0 ;
 5while (1<<(k+1)<=r-l+1) k++;
 6return _min(dp[l][k],dp[r-(1<<k)+1][k]);
 7}
 8int main()
 9{
10    #ifndef  ONLINE_JUDGE 
11	freopen("code/in.txt","r",stdin);
12  #endif
 1	ms(in,0);
 2	scanf("%d %d",&n,&m);
 3	for ( int i = 1 ; i <= n ; i++) edge[i].clear();
 4	for ( int i = 1 ; i <= m ; i++)
 5	{
 6	    int u,v,w;
 7	    char dir[5];
 8	    scanf("%d%d%d%s",&u,&v,&w,dir);
 9	    edge[u].push_back(make_pair(v,w));
10	    edge[v].push_back(make_pair(u,w));
11	}
1	p = 0 ;
2	dfs(1,0,0,-1);
3	rmq_init();
 1	scanf("%d",&q);
 2	while (q--)
 3	{
 4	    int u,v;
 5	    scanf("%d%d",&u,&v);
 6	   // cout<<"u:"<<u<<" v:"<<v<<endl;
 7	    int LCA = E[rmq_min(R[u],R[v])];
 8	//    cout<<"LCA:"<<LCA<<endl;
 9	    int ans;
10	    ans = dis[u]+dis[v]-2*dis[LCA];
11	    printf("%d\n",ans);
12	}
13  #ifndef ONLINE_JUDGE  
14  fclose(stdin);
15  #endif
16    return 0;
17}