HDU：2586-How far away

How far away

Time limit1000 ms

Memory limit32768 kB

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

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解题心得：

1. 题意就是给你一个双向图，每两点之间只有一条路径（其实就是一颗树），任意问你两点，要求输出两点间的最小距离。两种做法：
   • 第一种就是直接跑暴力的dfs，复杂度也就是n*m最大8e6，很简单就跑过了。
   • 第二种就是用LCA，两点之间最短的距离就是两点经过最近公共祖先形成的一条路。具体做法可以先预处理一下每个点到根节点的距离，存在数组dis里面，假如询问a，b之间的最小距离，就可以先得出最近公共祖先c然后答案就是dis[a] + dis[b] - 2*dis[c]。
2. 然后就是双向边怎么建成一棵树的问题，可以用邻接表存图，存双向边，在使用的时候任意选一个点为根节点，使用过的点就标记上，这样就可以形成一棵树，因为每两点之间只选择一个方向的边。

---

dfs跑暴力：

#include<stdio.h>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 4e4+100;
bool vis[maxn],flag;
int n,m,ans,a,b;
vector <pair<int,int> > ve[maxn];

void init()
{
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i=0;i<=n+1;i++)
        ve[i].clear();
    for(int i=1;i<n;i++)
    {
        int a,b,len;
        scanf("%d%d%d",&a,&b,&len);
        //邻接表存图
        ve[a].push_back(make_pair(b,len));
        ve[b].push_back(make_pair(a,len));
    }
}

void dfs(int pos,int sum_len)
{
    if(vis[pos] || flag)
        return ;
    vis[pos] = true;
    if(pos == b)
    {
        printf("%d\n",sum_len);
        flag = true;
        return;
    }
    for(int i=0;i<ve[pos].size();i++)
    {
        pair<int,int> p;
        p = ve[pos][i];
        dfs(p.first,sum_len + p.second);
    }
}

void solve()
{
    while(m--)
    {
        flag = false;
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&a,&b);
        dfs(a,0);
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        solve();
    }
}

---

LCA（tarjan写法）

#include<bits/stdc++.h>
using namespace std;
const int maxn = 4e4+100;
vector <pair<int,int> > ve[maxn];
int vis[maxn],n,father[maxn],sum[maxn],m,a,b;
bool flag;

void init()
{
    memset(father,0,sizeof(father));
    memset(vis,0,sizeof(vis));
    memset(sum,0,sizeof(sum));
    for(int i=0; i<=n; i++)
        ve[i].clear();
    for(int i=1; i<n; i++)
    {
        int a,b,len;
        scanf("%d%d%d",&a,&b,&len);
        ve[a].push_back(make_pair(b,len));
        ve[b].push_back(make_pair(a,len));
    }
}

void dfs(int pos,int len)
{
    vis[pos] = true;
    sum[pos] = len;
    for(int i=0; i<ve[pos].size(); i++)
    {
        int v = ve[pos][i].first;
        if(!vis[v])
            dfs(v,len+ve[pos][i].second);
    }
    return ;
}

int find(int x)
{
    if(x == father[x])
        return x;
    return father[x] = find(father[x]);
}

void tarjan(int pos)//跑tarjan
{
    vis[pos] = true;
    father[pos] = pos;
    if(flag)
        return;
    for(int i=0; i<ve[pos].size(); i++)
    {
        int v = ve[pos][i].first;
        if(!vis[v])
        {
            tarjan(v);
            father[v] = pos;
        }
        if(flag)
            return ;
    }
    if(pos == a || pos == b)
    {
        if(pos != a)
            swap(a,b);
        if(father[b])
        {
            int ans = find(father[b]);
            printf("%d\n",abs(sum[a]+sum[b]-sum[ans]*2));
            flag = true;
        }
    }
}

void query()
{
    while(m--)
    {
        flag = false;
        memset(father,0,sizeof(father));
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&a,&b);
        tarjan(1);
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        init();
        dfs(1,0);//先dfs预处理
        query();
    }
}