hdu 2586 How far away ？ 带权lca

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

思路：在lca的基础上加dis（距离根的距离）数组，在dfs处理好，ans=dis[a]-2*dis[LCA(a,b)]+dis[b]；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
#define true ture
#define false flase
using namespace std;
#define ll long long
#define inf 0xfffffff
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
#define maxn 40010
#define M 22
struct is
{
    int v,next,w;
} edge[maxn*2];
int deep[maxn],jiedge;
int dis[maxn];
int head[maxn];
int rudu[maxn];
int fa[maxn][M];
void add(int u,int v,int w)
{
    jiedge++;
    edge[jiedge].v=v;
    edge[jiedge].w=w;
    edge[jiedge].next=head[u];
    head[u]=jiedge;
}
void dfs(int u)
{
    for(int i=head[u]; i; i=edge[i].next)
    {
        int v=edge[i].v;
        int w=edge[i].w;
        if(!deep[v])
        {
            dis[v]=dis[u]+edge[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 )  // 注意此处，动手模拟一下，就会明白的
        {
            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(head,0,sizeof(head));
    memset(fa,0,sizeof(fa));
    memset(rudu,0,sizeof(rudu));
    memset(deep,0,sizeof(deep));
    jiedge=0;
}
int main()
{
    int x,n;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d%d",&n,&x);
        for(int i=1; i<n; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            rudu[v]++;
        }
        for(int i=1;i<=n;i++)
        {
            if(!rudu[i])
            {
                deep[i]=1;
                dis[i]=0;
                dfs(i);
                break;
            }
        }
        st(n);
        while(x--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d
",dis[a]-2*dis[LCA(a,b)]+dis[b]);
        }
    }
    return 0;
}