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}