HDU 2586 How Far Away? LCA水题

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

 

 

 

 

题意：给出一棵树，m个询问

每次询问给出a,b

问a,b的距离。

 

siz[i]  表示i到根节点的距离

 

思路：先求出lca=LCA(a,b)

 

则距离 = siz[a]+siz[b]-2*siz[lca]

 

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=40000+5;

struct Edge
{
    int to,next,w;
}edge[maxn<<1];
int head[maxn];
int tot;
int p[maxn][24];
int siz[maxn];
int dep[maxn];

void init(int n)
{
    tot=1;
    memset(head,-1,sizeof(head));
    memset(siz,0,sizeof(siz));
    memset(dep,0,sizeof(dep));

    for(int i=1;i<=n;i++)
        for(int j=0;j<24;j++)
            p[i][j]=-1;
}

void addedge(int u,int v,int w)
{
    edge[tot].to=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dfs(int n,int u,int fa)
{
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        int w=edge[i].w;
        if(!dep[v]&&v!=fa)
        {
            dep[v]=dep[u]+1;
            p[v][0]=u;
            siz[v]=siz[u]+w;
            dfs(n,v,u);
        }
    }
}

void init_lca(int n)
{
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            if(p[i][j-1]!=-1)
            {
                p[i][j]=p[p[i][j-1]][j-1];
            }
        }
    }
}

int solve(int n,int u,int v)
{
    if(u==v)
        return 0;
    if(dep[u]<dep[v])
    {
        swap(u,v);
    }

    int init_u=u;
    int init_v=v;

    int cnt;
    for(cnt=0;(1<<cnt)<=dep[u];cnt++)
        ;
    cnt--;

    for(int j=cnt;j>=0;j--)
    {
        if(dep[u]-(1<<j)>=dep[v])
            u=p[u][j];
    }

    if(u==v)
        return (siz[init_u]-siz[u]);

    for(int j=cnt;j>=0;j--)
    {
        if(p[u][j]!=-1&&p[u][j]!=p[v][j])
        {
            u=p[u][j];
            v=p[v][j];
        }
    }
    int lca=p[u][0];

    return (siz[init_u]+siz[init_v]-2*siz[lca]);

}

int main()
{
    int test;
    scanf("%d",&test);

    while(test--)
    {
        int n,m;
        scanf("%d%d",&n,&m);

        init(n);

        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }

        dfs(n,1,-1);
        init_lca(n);

        /*
        for(int i=1;i<=n;i++)
            printf("%d
",siz[i]);
        */

        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            //printf("eee
");
            printf("%d
",solve(n,u,v));
        }
    }

    return 0;
}