[bzoj1787][Ahoi2008]Meet 紧急集合(lca)

传送门

 

可以看出,三个点两两之间的lca会有一对相同,而另一个lca就是聚集点。

然后搞搞就可以求出距离了。

 

——代码

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #define MAXN 1000001
 5 
 6 using namespace std;
 7 
 8 int n, m, cnt, ans;
 9 int head[MAXN], to[MAXN], next[MAXN], deep[MAXN], f[MAXN][21];
10 
11 inline void add(int x, int y)
12 {
13     to[cnt] = y;
14     next[cnt] = head[x];
15     head[x] = cnt++;
16 }
17 
18 inline void dfs(int u)
19 {
20     int i, v;
21     deep[u] = deep[f[u][0]] + 1;
22     for(i = 0; f[u][i]; i++) f[u][i + 1] = f[f[u][i]][i];
23     for(i = head[u]; i != -1; i = next[i])
24     {
25         v = to[i];
26         if(!deep[v]) f[v][0] = u, dfs(v);
27     }
28 }
29 
30 inline int lca(int x, int y)
31 {
32     int i;
33     if(deep[x] < deep[y]) swap(x, y);
34     for(i = 20; i >= 0; i--)
35         if(deep[f[x][i]] >= deep[y])
36             x = f[x][i];
37     if(x == y) return x;
38     for(i = 20; i >= 0; i--)
39         if(f[x][i] != f[y][i])
40             x = f[x][i], y = f[y][i];
41     return f[x][0];
42 }
43 
44 int main()
45 {
46     int i, x, y, z, a;
47     scanf("%d %d", &n, &m);
48     memset(head, -1, sizeof(head));
49     for(i = 1; i < n; i++)
50     {
51         scanf("%d %d", &x, &y);
52         add(x, y);
53         add(y, x);
54     }
55     dfs(1);
56     for(i = 1; i <= m; i++)
57     {
58         scanf("%d %d %d", &x, &y, &z);
59         a = lca(x, y) ^ lca(x, z) ^ lca(y, z);
60         ans = deep[x] + deep[a] - 2 * deep[lca(x, a)];
61         ans += deep[y] + deep[a] - 2 * deep[lca(y, a)];
62         ans += deep[z] + deep[a] - 2 * deep[lca(z, a)];
63         printf("%d %d\n", a, ans);
64     }
65     return 0;
66 }
View Code

 

posted @ 2017-05-04 19:03  zht467  阅读(79)  评论(0编辑  收藏  举报