HDU 2586 倍增法求lca

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

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <map>
 #define ll  __int64
 #define mod 1000000007
 #define dazhi 2147483647
 #define bug() printf("!!!!!!!")
 using namespace  std;
 #define maxn 40010
 #define M 22
 struct node
 {
     int pre;
     int to;
     int w;
 }N[maxn*];
 int pre[maxn];
 int deep[maxn],nedge=;
 int dis[maxn];
 int rudu[maxn];
 int fa[maxn][M];
 int t;
 int f1,t1,w1;
 int l,r;
 int n,m;
 void add(int from ,int to,int ww)
 {
     nedge++;
     N[nedge].to=to;
     N[nedge].pre=pre[from];
     N[nedge].w=ww;
     pre[from]=nedge;
 }
 void dfs(int u)
 {
     for(int i=pre[u];i;i=N[i].pre)
     {
         int v=N[i].to;
         if(deep[v]==)
         {
             dis[v]=dis[u]+N[i].w;
             deep[v]=deep[u]+;
             fa[v][]=u;
             dfs(v);
         }
     }
 }
 void st(int n)
 {
     for(int j=;j<M;j++)
         for(int i=;i<=n;i++)
          fa[i][j]=fa[fa[i][j-]][j-];
 }
 int lca(int u,int v)
 {
     if(deep[u]<deep[v]) swap(u,v);
     int d=deep[u]-deep[v];
     int i;
     for(i=;i<M;i++)
     {
         if((<<i)&d)
         {
             u=fa[u][i];
         }
     }
     if(u==v) return u;
     for(i=M-;i>=;i--)
     {
         if(fa[u][i]!=fa[v][i])
         {
             u=fa[u][i];
             v=fa[v][i];
         }
     }
     u=fa[u][];
     return u;
 }
 void init()
 {
    memset(rudu,,sizeof(rudu));
    memset(dis,,sizeof(dis));
    memset(deep,,sizeof(deep));
    memset(fa,,sizeof(fa));
    memset(N,,sizeof(N));
    memset(pre,,sizeof(pre));
    nedge=;
 }
 int main()
 {
     while(scanf("%d",&t)!=EOF)
     {
         for(int i=;i<=t;i++)
         {
             init();
             scanf("%d %d",&n,&m);
             for(int j=;j<n;j++)
             {
                scanf("%d %d %d",&f1,&t1,&w1);
                add(f1,t1,w1);
                rudu[t1]++;
             }
             for(int j=;j<=n;j++)
             {
                 if(rudu[j]==)
                 {
                     deep[j]=;
                     dis[j]=;
                     dfs(j);
                     break;
                 }
             }
             st(n);
             for(int j=;j<=m;j++)
             {
                int aa,bb;
                scanf("%d %d",&aa,&bb);
                printf("%d\n",dis[aa]+dis[bb]-*dis[lca(aa,bb)]);
             }
         }
     }
     return ;
 }