练习小机房的树

Posted 2021-01-12 hwjia

2370 小机房的树

//讲真我觉得这题面有点恶心QAQ

大体思路：树上的最短路径一定经过两点lca， 预处理一个dis数组， 存每个点到根节点的距离， 两点（x， y） 之间的距离即为dis[x] + dis[y] - 2*dis[lca] ；

然后倍增求lca即可，dis数组在预处理每个点深度的时候一并处理

日常，“妙啊”

代码君qwq

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<iostream>
 4 using namespace std;
 5 const int maxn = 500050, maxm = 500050;
 6 int n, m, s, num = 0, log2n;
 7 int head[maxm], jump[maxn][23], dep[maxn], dis[maxn];
 8 struct edge {
 9     int nxt, to, dis;
10 }e[maxm<<1];
11 void add(int from, int to, int dis) {
12     e[++num].nxt = head[from];
13     e[num].to = to;
14     e[num].dis = dis;
15     head[from] = num;
16 }
17 void dfs(int v) {
18     for(int i = head[v]; i; i = e[i].nxt) {
19         int u = e[i].to;
20         if(jump[v][0] != u) {
21             jump[u][0] = v;
22             dep[u] = dep[v] + 1;
23             dis[u] = dis[v]+e[i].dis;
24             dfs(u);
25         }
26     }
27 }
28 void init() {
29 /**/for(int i = 1; i <= log2n; i++) 
30         for(int j = 1; j <= n; j++) 
31             jump[j][i] = jump[jump[j][i-1]][i-1];
32 }
33 int LCA(int x, int y) {
34     if(dep[x] < dep[y]) swap(x, y);
35     int t = dep[x] - dep[y];
36     for(int i = 0; i <= log2n; i++) {
37         if(t&(1<<i)) x = jump[x][i];
38     }
39     if(x == y) return x;
40     for(int i = log2n; i >= 0; i--) {
41         if(jump[x][i] != jump[y][i]) {
42             x = jump[x][i];
43             y = jump[y][i];
44         }
45     }
46     return jump[x][0];
47 }
48 int main() {
49     scanf("%d", &n);
50     for(int i = 1; i < n; i++) {
51         int x, y, z;
52         scanf("%d%d%d", &x, &y, &z);
53         add(x, y, z), add(y, x, z);
54     }    
55     s = 0;
56     log2n = log(n)/log(2)+1;
57     dep[s] = 1;
58     jump[s][0] = 0;
59     dis[s] = 0;
60     dfs(s);
61     init();
62     scanf("%d", &m);
63     for(int i = 1; i <= m; i++) {
64         int x, y, ans = 0, lca;
65         scanf("%d%d", &x, &y);
66         lca = LCA(x, y);
67         ans = dis[x] + dis[y] - 2*dis[lca];
68         printf("%d
", ans);
69     }
70     return 0;
71 }