Description
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
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
2 5
4 1
6 0
思路：唔，直接暴力求出三个lca判断一下哪个lca是题目所要求的点就好了
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #define maxn 1000009 6 #define D 20 7 using namespace std; 8 int head[maxn],nex[maxn],point[maxn]; 9 int fa[maxn][D+10],x,y,z,now,deep[maxn]; 10 void add(int x,int y) 11 { 12 nex[++now] = head[x]; 13 head[x] = now; 14 point[now] = y; 15 } 16 void bfs(int s) 17 { 18 queue<int>q; 19 q.push(s); 20 while(!q.empty()) 21 { 22 int u = q.front(); 23 //printf("%d ",u); 24 q.pop(); 25 for(int i = head[u]; i; i=nex[i]) 26 { 27 int k = point[i]; 28 if(k == fa[u][0])continue; 29 q.push(k); 30 fa[k][0] = u; 31 for(int j = 1; j < D; j++)fa[k][j] = fa[fa[k][j-1]][j-1]; 32 deep[k] = deep[u] + 1; 33 } 34 } 35 } 36 int lca(int x,int y) 37 { 38 if(deep[x]>deep[y])swap(x,y); 39 int tx = x, ty = y,delta = deep[y] - deep[x]; 40 for(int i = 0;delta;delta>>=1,i++)if(delta&1) 41 { 42 ty = fa[ty][i]; 43 } 44 if(tx == ty)return tx; 45 for(int i = D; i>=0;i--) 46 { 47 if(fa[tx][i] == fa[ty][i])continue; 48 tx = fa[tx][i]; 49 ty = fa[ty][i]; 50 } 51 return fa[tx][0]; 52 } 53 54 int main() 55 { 56 int n,m; 57 scanf("%d%d",&n,&m); 58 for(int i=1;i<n;i++) 59 { 60 scanf("%d%d",&x,&y); 61 add(x,y); 62 add(y,x); 63 } 64 //puts("0"); 65 bfs(1); 66 //puts("1"); 67 for(int i = 1; i <= m; i++) 68 { 69 scanf("%d%d%d",&x,&y,&z); 70 int a = lca(x,y),b=lca(x,z),c= lca(y,z); 71 int ans = (deep[x]+deep[y]-deep[a])+deep[z]-2*deep[lca(a,z)],ansj = a; 72 int u; 73 if((u = (deep[x]+deep[z]-deep[b]+deep[y]-2*deep[lca(b,y)]))<ans) 74 { 75 ans = u; 76 ansj = b; 77 } 78 if((u = (deep[y]+deep[z]-deep[c]+deep[x]-2*deep[lca(c,x)]))<ans) 79 { 80 ans = u; 81 ansj = c; 82 } 83 printf("%d %d ",ansj,ans); 84 } 85 return 0; 86 }