How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14685    Accepted Submission(s): 5554


Problem Description
There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this "How far is it if I want to go from house A to house B"? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path("simple" means you can't visit a place twice) between every two houses. Yout task is to answer all these curious people.
 

 

Input
First line is a single integer T(T<=10), indicating the number of test cases.
  For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0<k<=40000).The houses are labeled from 1 to n.
  Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
 

 

Output
For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.
 

 

Sample Input
2 3 2 1 2 10 3 1 15 1 2 2 3 2 2 1 2 100 1 2 2 1
 

 

Sample Output
10 25 100 100
 

 

Source
ECJTU 2009 Spring Contest
 题意:给你一颗带权树 求任意两个结点间的最短距离
 题解:dis存结点到根的距离+lca   dis=dis[a]+dis[b]-2*dis[lca(a,b)]; 倍增法求lca
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <algorithm>
  6 #include <stack>
  7 #include <queue>
  8 #include <cmath>
  9 #include <map>
 10 #define ll  __int64
 11 #define mod 1000000007
 12 #define dazhi 2147483647
 13 #define bug() printf("!!!!!!!")
 14 using namespace  std;
 15 #define maxn 40010
 16 #define M 22
 17 struct node
 18 {
 19     int pre;
 20     int to;
 21     int w;
 22 }N[maxn*2];
 23 int pre[maxn];
 24 int deep[maxn],nedge=0;
 25 int dis[maxn];
 26 int rudu[maxn];
 27 int fa[maxn][M];
 28 int t;
 29 int f1,t1,w1;
 30 int l,r;
 31 int n,m;
 32 void add(int from ,int to,int ww)
 33 {
 34     nedge++;
 35     N[nedge].to=to;
 36     N[nedge].pre=pre[from];
 37     N[nedge].w=ww;
 38     pre[from]=nedge;
 39 }
 40 void dfs(int u)
 41 {
 42     for(int i=pre[u];i;i=N[i].pre)
 43     {
 44         int v=N[i].to;
 45         if(deep[v]==0)
 46         {
 47             dis[v]=dis[u]+N[i].w;
 48             deep[v]=deep[u]+1;
 49             fa[v][0]=u;
 50             dfs(v);
 51         }
 52     }
 53 }
 54 void st(int n)
 55 {
 56     for(int j=1;j<M;j++)
 57         for(int i=1;i<=n;i++)
 58          fa[i][j]=fa[fa[i][j-1]][j-1];
 59 }
 60 int lca(int u,int v)
 61 {
 62     if(deep[u]<deep[v]) swap(u,v);
 63     int d=deep[u]-deep[v];
 64     int i;
 65     for(i=0;i<M;i++)
 66     {
 67         if((1<<i)&d)
 68         {
 69             u=fa[u][i];
 70         }
 71     }
 72     if(u==v) return u;
 73     for(i=M-1;i>=0;i--)
 74     {
 75         if(fa[u][i]!=fa[v][i])
 76         {
 77             u=fa[u][i];
 78             v=fa[v][i];
 79         }
 80     }
 81     u=fa[u][0];
 82     return u;
 83 }
 84 void init()
 85 {
 86    memset(rudu,0,sizeof(rudu));
 87    memset(dis,0,sizeof(dis));
 88    memset(deep,0,sizeof(deep));
 89    memset(fa,0,sizeof(fa));
 90    memset(N,0,sizeof(N));
 91    memset(pre,0,sizeof(pre));
 92    nedge=0;
 93 }
 94 int main()
 95 {
 96     while(scanf("%d",&t)!=EOF)
 97     {
 98         for(int i=1;i<=t;i++)
 99         {
100             init();
101             scanf("%d %d",&n,&m);
102             for(int j=1;j<n;j++)
103             {
104                scanf("%d %d %d",&f1,&t1,&w1);
105                add(f1,t1,w1);
106                rudu[t1]++;
107             }
108             for(int j=1;j<=n;j++)
109             {
110                 if(rudu[j]==0)
111                 {
112                     deep[j]=1;
113                     dis[j]=0;
114                     dfs(j);
115                     break;
116                 }
117             }
118             st(n);
119             for(int j=1;j<=m;j++)
120             {
121                int aa,bb;
122                scanf("%d %d",&aa,&bb);
123                printf("%d\n",dis[aa]+dis[bb]-2*dis[lca(aa,bb)]);
124             }
125         }
126     }
127     return 0;
128 }