One of the four solution LCA: multiplication and examples

hdu-2586  http://acm.hdu.edu.cn/showproblem.php?pid=2586

// meaning of the questions: a village there are n houses, only one path between two houses, and the village is China Unicom. Q m queries distance between the two houses, the answer to the query output every time. n (2 <= n <= 40000) m (1 <= m <= 200)

. 1 #include <cstdio>
 2 #include <CString>
 . 3 #include <algorithm>
 . 4  the using  namespace STD;
 . 5  
. 6  const  int MAXN = 4e4 + . 5 ;
 . 7  
. 8  struct Node { // V will be appreciated that for the current node, and u can be is understood to father node 
. 9      int U, v, W, Next; // W is a two u-> v is from 
10 } Edge [ 2 * MAXN]; // undirected graph for a 2 * 
. 11  
12 is  int n-, m;
 13 is  int NUM = . 1 ; // number of nodes 
14 int head [ 2 * MAXN]; // used to map a node ID value 
15  int Deep [MAXN]; // a point in the depth (tree) 
16  int F [MAXN] [ 21 is ]; // F [ i] [J] denotes the i 2 ^ j ancestors 
. 17  int G [MAXN]; // save distance from a point to the root of 
18 is  
. 19  void add_edge member ( int X, int Y, int Z) { // contribution 
20 is      Edge [NUM] .u = X;
 21 is      Edge [NUM] .v = Y;
 22 is      Edge [NUM] .W = Z;
 23 is     Edge [NUM] = .next head [X];
 24      head [X] NUM = ++ ;
 25  }
 26 is  
27  void DFS ( int CUR) { // calculates each node's father node number, depth, and the roots distance 
28      for ( int I = head [CUR]; I = -! . 1 ; I = Edge [I] .next) {
 29          IF (Deep [Edge [I] .v] == 0 ) {
 30              Deep [Edge [I] .v] = Deep [CUR] + . 1 ;
 31 is              F [Edge [I] .v] [ 0 ] = CUR;
 32              G [Edge [I] .v] = G [CUR] + Edge [I] .w;
33             dfs(edge[i].v);
34         }
35     }
36 }
37 
38 void PRE() {// 倍增求f[i][j]
39     for(int j = 1; j <= 19; ++j) {
40         for(int i = 1; i <= n; ++i) {
41             f[i][j] = f[f[i][j-1]][j-1];
42         }
43     }
44 }
45 
46 int LCA(int x, int y) {// 倍增思想
47     if(deep[x] < deep[y]) swap(x, y);
48     for(int i = 19; i >= 0; --i) {
49         if(deep[f[x][i]] >= deep[y]) {
50             x = f[x][i];
51         }
52     }
53     if(x == y) return x;
54     for(int i = 19; i >= 0; --i) {
55         if(f[x][i] != f[y][i]) {
56             x = f[x][i];
57             y = f[y][i];
58         }
59     }
60     return f[x][0];
61 }
62 
63 int main() {
64     int T;
65     scanf("%d", &T);
66     while(T--) {
67         memset(head, -1, sizeof(head));
68         memset(deep, 0, sizeof(deep));
69         memset(f, 0, sizeof(f));
70         memset(g, 0, sizeof(g));
71 
72         scanf("%d%d", &n, &m);
73         int s, e, v;
74         for(int i = 0; i != n-1; ++i) {
75             scanf("%d%d%d", &s, &e, &v);
76             add_edge(s, e, v);
77             add_edge(e, s, v);
78         }
79         deep[1] = 1;
80         dfs(1);
81         PRE();
82         for(int i = 0; i != m; ++i) {
83             scanf("%d%d", &s, &e);
84             printf("%d\n", g[s]+g[e]-2*g[LCA(s, e)]);// 画图即可理解
85         }
86     }
87     return 0;
88 }