LCA 板子题

来自某校内赛（2333

题目大意：

    给你一棵树，并有m次询问。每次给出A,B,C三个数，然后要求一个点x使得A\B\C三点路径最短。

思路：

  这一看就是极好的题，ZZ知道直接倍增求LCA好了。其中三个LCA中，不相等的一个必然为正解。（然后我就一直挂在倍增求LCA了2333

代码：（来自std）

#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define N 600010#define lowbit(x) (x & (-x))int to[N * 2], next[N * 2], p[N], q[N], fa[N], d[N], f[N][20], n, m, i, j, ed,tot;int gi() {     int s = 0; char c = getchar();     while (c < '0' || c > '9') c = getchar();     while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();     return s; } void add(int u, int v) {     to[++tot] = v, next[tot] = p[u], p[u] = tot; } void bfs() {     int s = 0, t = 0;     q[++t] = 1, d[1] = 1;     while (s < t) {         int u = q[++s], i;         for (i = p[u]; i != -1; i = next[i])             if (to[i] != fa[u])                 q[++t] = to[i], fa[to[i]] = u, d[to[i]] = d[u] + 1;     }     for (i = 1; i <= n; i++) f[i][0] = fa[i];     for (i = 1; i <= 17; i++)         for (j = 1; j <= n; j++)             f[j][i] = f[f[j][i - 1]][i - 1]; }int Lca(int u, int v) {     if (d[u] < d[v]) swap(u, v);     for (int i = 17; i >= 0; i--)         if (d[f[u][i]] >= d[v]) u = f[u][i];     if (u == v) return u;     for (int i = 17; i >= 0; i--)         if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];     return fa[u]; } int main(){   //freopen("xian.in","r",stdin); freopen("xian.out","w",stdout);   memset(p,-1,sizeof(p));   scanf("%d%d",&n,&m);   for(int i=1;i<n;i++){     int u=gi(),v=gi();     add(u,v); add(v,u);   }   bfs();   for (i = 1; i <= m; i++) {         int x = gi(), y = gi(), z = gi();         int a = Lca(x, y), b = Lca(x, z), c = Lca(y, z); int rt = Lca(a, z);         int d1 = d[x] + d[y] - 2 * d[a] + d[a] - d[rt] + d[z] - d[rt];         int d2 = d[x] + d[z] - 2 * d[b] + d[b] - d[rt] + d[y] - d[rt];         int d3 = d[y] + d[z] - 2 * d[c] + d[c] - d[rt] + d[x] - d[rt];         int D = d1, g = a;         if (d2 < D) D = d2, g = b;         if (d3 < D) D = d3, g = c; //进行比较，其实按我的思路也是可以的        printf("%d %d\n", g, D);     }     return 0;}

总结：

  1.要背好板子

 2.板子不能背错（痛苦

 3.不要相信学长