1787: [Ahoi2008]Meet 紧急集合

Time Limit: 20 Sec  Memory Limit: 162 MB
Submit: 4669  Solved: 2246
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Description

Input

Output

Sample Input

6 4
1 2
2 3
2 4
4 5
5 6
4 5 6
6 3 1
2 4 4
6 6 6

Sample Output


5 2
2 5
4 1
6 0

HINT

 

LCA果然还是玄学，倍增父节点这点理解好久

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 
 5 const int MAXN=1000005;
 6 struct Edge
 7 {
 8     int to,next;
 9 }E[MAXN];
10 int node,head[MAXN];
11 int deep[MAXN],fa[MAXN][20];
12 bool vis[MAXN];
13 int n,m;
14 
15 void insert(int u,int v)
16 {
17     E[++node]=(Edge){v,head[u]};head[u]=node;
18     E[++node]=(Edge){u,head[v]};head[v]=node;
19 }
20 
21 void dfs(int x)
22 {
23     vis[x]=1;
24     for(int i=1;i<=18;i++)
25     {
26         if(deep[x]<(1<<i)) break;
27         fa[x][i]=fa[fa[x][i-1]][i-1];
28     }
29     for(int i=head[x];i;i=E[i].next)
30     {
31         if(vis[E[i].to]) continue;
32         deep[E[i].to]=deep[x]+1;
33         fa[E[i].to][0]=x;
34         dfs(E[i].to);
35     }
36 }
37 
38 int lca(int x,int y)
39 {
40     if(deep[x]<deep[y]) swap(x,y);
41     int d=deep[x]-deep[y];
42     for(int i=0;i<=18;i++)
43         if((1<<i)&d) x=fa[x][i];
44     for(int i=18;i>=0;i--)
45         if(fa[x][i]!=fa[y][i])
46         {
47             x=fa[x][i];
48             y=fa[y][i];
49         }
50     if(x==y) return x;
51     else return fa[x][0];
52 }
53 
54 int dis(int x,int y)
55 {
56     int t=lca(x,y);
57     return deep[x]+deep[y]-2*deep[t];
58 }
59 
60 int main()
61 {
62     scanf("%d%d",&n,&m);
63     for(int i=1;i<n;i++)
64     {
65         int u,v;
66         scanf("%d%d",&u,&v);
67         insert(u,v);
68     }
69     dfs(1);
70     for(int i=1;i<=m;i++)
71     {
72         int x,y,z;
73         scanf("%d%d%d",&x,&y,&z);
74         int a=lca(x,y),b=lca(x,z),c=lca(y,z),t;
75         if(a==b) t=c;
76         else if(b==c) t=a;
77         else if(a==c) t=b;
78         int ans=dis(x,t)+dis(y,t)+dis(z,t);
79         printf("%d %d\n",t,ans);
80     }
81     return 0;
82 }

 

posted @ 2019-04-02 15:58  InWILL  阅读(270)  评论(0)  编辑  收藏  举报