题目链接：

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

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

100

题意：

给一个无向图,问两点i和j之间的距离是多少;

思路：

由于询问规模大,所以就无向图转化成树,然后寻找lca,然后再算距离;

把lca转化成RMQ问题,我是用的线段树找的RMQ;

AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N=4e4+;

typedef long long ll;

const double PI=acos(-1.0);

int n,m,t,dis[N],dep[N],vis[N],w[N];

vector<int>ve[N];

queue<int>qu;

int u[N],v[N],d[N],fa[N],a[*N],tot,first[N];

struct Tree

{

    int l,r,ans;

};

Tree tree[*N];

void bfs()//bfs把图转化成树;

{

    qu.push();

    vis[]=;

    dep[]=;

    while(!qu.empty())

    {

        int top=qu.front();

        qu.pop();

        int len=ve[top].size();

        for(int i=;i<len;i++)

        {

            int x=ve[top][i];

            if(!vis[x])

            {

                vis[x]=;

                qu.push(x);

                dep[x]=dep[top]+;

                fa[x]=top;

            }

        }

    }

}

int dfs(int num)//dfs找到lca的数组,并计算i到1的dis

{

    vis[num]=;

    first[num]=tot;

    a[tot++]=num;

    int len=ve[num].size();

    for(int i=;i<len;i++)

    {

        int x=ve[num][i];

        if(vis[x])

        {

            dis[x]=dis[num]+d[x];

            dfs(x);

            a[tot++]=num;

        }

    }

}

void Pushup(int node)

{

    if(dep[a[tree[*node].ans]]>=dep[a[tree[*node+].ans]])tree[node].ans=tree[*node+].ans;

    else tree[node].ans=tree[*node].ans;

}//tree[node].ans为数组里的lca的位置;

void build(int node,int L,int R)

{

    tree[node].l=L;

    tree[node].r=R;

    if(L>=R)

    {

        tree[node].ans=L;

        return ;

    }

    int mid=(L+R)>>;

    build(*node,L,mid);

    build(*node+,mid+,R);

    Pushup(node);

}

int query(int node,int L,int R)

{

    if(L<=tree[node].l&&R>=tree[node].r)return tree[node].ans;

    int mid=(tree[node].l+tree[node].r)>>;

    if(R<=mid)return query(*node,L,R);

    else if(L>mid)return query(*node+,L,R);

    else

    {

        int pos1=query(*node,L,mid),pos2=query(*node+,mid+,R);

        if(dep[a[pos1]]>=dep[a[pos2]])return pos2;

        else return pos1;

    }

}

int main()

{

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d",&n,&m);

        for(int i=;i<=n;i++)

        {

            ve[i].clear();

        }

        for(int i=;i<n;i++)

        {

            scanf("%d%d%d",&u[i],&v[i],&w[i]);

            ve[u[i]].push_back(v[i]);

            ve[v[i]].push_back(u[i]);

        }

        memset(vis,,sizeof(vis));

        bfs();

        for(int i=;i<n;i++)

        {

            if(fa[u[i]]==v[i])

            {

                d[u[i]]=w[i];

            }

            else

            {

                d[v[i]]=w[i];

            }

        }

        tot=;

        dfs();

        build(,,tot-);

        int ql,qr;

        for(int i=;i<m;i++)

        {

            scanf("%d%d",&ql,&qr);

            int pos;

            if(first[ql]<=first[qr])

            pos=query(,first[ql],first[qr]);

            else pos=query(,first[qr],first[ql]);

            printf("%d\n",dis[ql]-dis[a[pos]]-dis[a[pos]]+dis[qr]);

        }

    }

    return ;

}

巴特西

hdu-2586 How far away ？(lca+bfs+dfs+线段树)

How far away ？

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