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*2];
int pre[maxn];
int deep[maxn],nedge=0;
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]==0)
        {
            dis[v]=dis[u]+N[i].w;
            deep[v]=deep[u]+1;
            fa[v][0]=u;
            dfs(v);
        }
    }
}
void st(int n)
{
    for(int j=1;j<M;j++)
        for(int i=1;i<=n;i++)
         fa[i][j]=fa[fa[i][j-1]][j-1];
}
int lca(int u,int v)
{
    if(deep[u]<deep[v]) swap(u,v);
    int d=deep[u]-deep[v];
    int i;
    for(i=0;i<M;i++)
    {
        if((1<<i)&d)
        {
            u=fa[u][i];
        }
    }
    if(u==v) return u;
    for(i=M-1;i>=0;i--)
    {
        if(fa[u][i]!=fa[v][i])
        {
            u=fa[u][i];
            v=fa[v][i];
        }
    }
    u=fa[u][0];
    return u;
}
void init()
{
   memset(rudu,0,sizeof(rudu));
   memset(dis,0,sizeof(dis));
   memset(deep,0,sizeof(deep));
   memset(fa,0,sizeof(fa));
   memset(N,0,sizeof(N));
   memset(pre,0,sizeof(pre));
   nedge=0;
}
int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        for(int i=1;i<=t;i++)
        {
            init();
            scanf("%d %d",&n,&m);
            for(int j=1;j<n;j++)
            {
               scanf("%d %d %d",&f1,&t1,&w1);
               add(f1,t1,w1);
               rudu[t1]++;
            }
            for(int j=1;j<=n;j++)
            {
                if(rudu[j]==0)
                {
                    deep[j]=1;
                    dis[j]=0;
                    dfs(j);
                    break;
                }
            }
            st(n);
            for(int j=1;j<=m;j++)
            {
               int aa,bb;
               scanf("%d %d",&aa,&bb);
               printf("%d
",dis[aa]+dis[bb]-2*dis[lca(aa,bb)]);
            }
        }
    }
    return 0;
}