Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other.
The boys decided to have fun and came up with a plan. Namely, in some day in the morning Misha will ride the underground from station s to station f by the shortest path, and will draw with aerosol an ugly text "Misha was here" on every station he will pass through (including s and f). After that on the same day at evening Grisha will ride from station t to station f by the shortest path and will count stations with Misha's text. After that at night the underground workers will wash the texts out, because the underground should be clean.
The boys have already chosen three stations a, b and c for each of several following days, one of them should be station s on that day, another should be station f, and the remaining should be station t. They became interested how they should choose these stations s, f, t so that the number Grisha will count is as large as possible. They asked you for help.
The first line contains two integers n and q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of stations and the number of days.
The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi ≤ n). The integer pi means that there is a route between stations pi and i. It is guaranteed that it's possible to reach every station from any other.
The next q lines contains three integers a, b and c each (1 ≤ a, b, c ≤ n) — the ids of stations chosen by boys for some day. Note that some of these ids could be same.
Print q lines. In the i-th of these lines print the maximum possible number Grisha can get counting when the stations s, t and f are chosen optimally from the three stations on the i-th day.
3 2
1 1
1 2 3
2 3 3
2
3
4 1
1 2 3
1 2 3
2
In the first example on the first day if s = 1, f = 2, t = 3, Misha would go on the route 1 
2, and Grisha would go on the route 3 1 2. He would see the text at the stations 1 and 2. On the second day, if s = 3, f = 2, t = 3, both boys would go on the route 3 1 2. Grisha would see the text at 3 stations.
In the second examle if s = 1, f = 3, t = 2, Misha would go on the route 1 2 3, and Grisha would go on the route 2 3 and would see the text at both stations.
树上a到b的最短路径就是 从a，b的LCA走向b的这条唯一路径。
对于本题，分别求出(a,b),(b,c),(c,a)的LCA 为x,y,z 不妨设其中深度最大的为x 即（a,b）的LCA。
若f为c，则此时的公共距离为从x到c的路径所经过的点数。
若f为a或b，不妨设为a，则此时的公共距离为 从x到a的路径所经过的点数 与从y到a的路径所经过的点数中的较小的一个。因为x为深度最大的，a点的深度为一定值（且a、x所求得的值就是深度差+1），故x出发的必然更小。
因此，只需要求两两LCA中深度最大的到三个点中最大的路径经过的点数即可。
1 #include <iostream> 2 #include <string> 3 #include <algorithm> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <queue> 8 #include <set> 9 #include <map> 10 #include <list> 11 #include <vector> 12 #include <stack> 13 #define mp make_pair 14 //#define P make_pair 15 #define MIN(a,b) (a>b?b:a) 16 //#define MAX(a,b) (a>b?a:b) 17 typedef long long ll; 18 typedef unsigned long long ull; 19 const int MAX=1e5+5; 20 const int INF=1e8+5; 21 using namespace std; 22 //const int MOD=1e9+7; 23 typedef pair<ll,int> pii; 24 const double eps=0.00000001; 25 const int MAX_V=1e5+5; 26 const int MAX_LOG_V=20; 27 vector<int> G[MAX_V];//图的邻接表表示 28 int root;//根节点编号 29 int parent[MAX_LOG_V][MAX_V];//向上走2^k步时到达的节点 （超过根时记为-1） 30 int depth[MAX_V];//节点的深度 31 void dfs(int v,int p,int d) 32 { 33 parent[0][v]=p; 34 depth[v]=d; 35 for(int i=0;i<G[v].size();i++) 36 { 37 if(G[v][i]!=p) 38 dfs(G[v][i],v,d+1); 39 } 40 } 41 //预处理 42 void init(int V) 43 { 44 //预处理出 parent[0]和depth 45 dfs(root,-1,0); 46 //预处理出parent 47 for(int k=0;k+1<MAX_LOG_V;k++) 48 for(int v=0;v<V;v++) 49 { 50 if(parent[k][v]<0) 51 parent[k+1][v]=-1; 52 else 53 parent[k+1][v]=parent[k][parent[k][v]]; 54 } 55 } 56 //计算u和v的LCA 57 int lca(int u,int v) 58 { 59 //让u和v向上走到同一深度 60 if(depth[u]>depth[v]) 61 swap(u,v); 62 for(int k=0;k<MAX_LOG_V;k++) 63 { 64 if(((depth[v]-depth[u])>>k)&1) 65 v=parent[k][v]; 66 } 67 if(u==v) 68 return u; 69 for(int k=MAX_LOG_V-1;k>=0;k--) 70 { 71 if(parent[k][u]!=parent[k][v]) 72 { 73 u=parent[k][u];v=parent[k][v]; 74 } 75 } 76 return u==v?u:parent[0][u]; 77 } 78 int dis(int x,int y) 79 { 80 if(depth[x]<depth[y]) 81 swap(x,y); 82 int lo=lca(x,y); 83 if(lo==y) 84 return depth[x]-depth[y]+1; 85 else 86 return depth[x]+depth[y]-2*depth[lo]+1; 87 } 88 int n,q; 89 int main() 90 { 91 scanf("%d%d",&n,&q); 92 for(int i=1;i<n;i++) 93 { 94 int x; 95 scanf("%d",&x);--x; 96 G[x].push_back(i); 97 } 98 init(n); 99 while(q--) 100 { 101 int a[3],b[3],c[3]; 102 for(int i=0;i<3;i++) 103 scanf("%d",&a[i]); 104 for(int i=0;i<3;i++) 105 --a[i]; 106 int an=-1,anz=0; 107 for(int i=0;i<3;i++) 108 { 109 int x=i,y=(i+2)%3; 110 b[i]=lca(a[x],a[y]); 111 if(depth[b[i]]>an) 112 { 113 an=depth[b[i]]; 114 anz=b[i]; 115 } 116 } 117 an=0; 118 for(int i=0;i<3;i++) 119 an=max(an,dis(a[i],anz)); 120 printf("%d ",an); 121 } 122 }