How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 34457    Accepted Submission(s): 13982

 

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
 
题意：给出房屋的个数 n ，边数为 n - 1，询问 m 次，每次询问两房屋之间的路径长度
 
思路：数据比较大，用 tarjin 离线求 lca，过程中维护距离值，dis[i]为点 i 到根节点1的距离
 

tarjin求lca

讲解：https://www.cnblogs.com/JVxie/p/4854719.html

  
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 4e4 + 10;

int t, n, m, u, v, w, head[N], head2[N], tot, tot2, fa[N], dis[N];
bool vis[N];
struct node
{
    int v, next, dis;
}edge[N << 1];

struct query
{
    int lca, u, v, next;
}q[205 << 1];

void init()
{
    tot = 0;
    tot2 = 0;
    memset(head, -1, sizeof(head));
    memset(head2, -1, sizeof(head2));
    memset(vis, 0, sizeof(vis));
    memset(dis, 0, sizeof(dis));
}

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

void add2(int u, int v)
{
    q[tot2].next = head2[u];
    q[tot2].u = u;
    q[tot2].v = v;
    head2[u] = tot2++;
}

int Find(int x)
{
    if(fa[x] != x)
        fa[x] = Find(fa[x]);
    return fa[x];
}

void tarjin(int u)
{
    fa[u] = u;
    vis[u] = 1;
    for(int i = head[u]; ~i; i = edge[i].next)
    {
        int v = edge[i].v, w = edge[i].dis;
        if(!vis[v])
        {
            dis[v] = dis[u] + w;
            tarjin(v);
            fa[v] = u;
        }
    }
    for(int i = head2[u]; ~i; i = q[i].next)
    {
        int v = q[i].v;
        if(vis[v])
        {
            q[i].lca = q[i ^ 1].lca = Find(v);
        }
    }
}

int main()
{
    scanf("%d", &t);
    while(t--)
    {
        init();
        scanf("%d%d", &n, &m);
        for(int i = 1; i < n; ++i)
        {
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
            add(v, u, w);
        }
        for(int i = 1; i <= m; ++i)
        {
            scanf("%d%d", &u, &v);
            add2(u, v);
            add2(v, u);
        }
        tarjin(1);
        for(int i = 0; i < m; ++i)
        {
            int u = q[2 * i].u;
            int v = q[2 * i].v;
            int w = q[2 * i].lca;
            cout<<dis[u] + dis[v] - 2 * dis[w]<<'\n';
        }
    }
    return 0;
}

 

树上倍增

树上倍增讲解和用法：https://blog.csdn.net/saramanda/article/details/54963914

树上倍增求lca：https://www.cnblogs.com/lbssxz/p/11114819.html

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int N = 4e4 + 10;

//dis[i]:点i到根节点的距离
//fa[i][j]:节点i的第2^j个父亲，每个点最多2^(logN)个父亲，所以第二维开logN
//dep[i]:当前节点的深度
int n, m, dis[N], head[N], fa[N][21], tot, dep[N];
struct Edge
{
    int to, next, w;
}edge[N << 1];

void init()
{
    tot = 0;
    memset(dis, 0, sizeof(dis));
    memset(head, -1, sizeof(head));
    memset(fa, 0, sizeof(fa));
}

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

void dfs(int u, int father)   //当前节点和它的父亲
{
    dep[u] = dep[father] + 1;
    for(int i = 1; (1 << i) <= dep[u]; ++i)
    {
        if(fa[u][i - 1])
            fa[u][i] = fa[fa[u][i - 1]][i - 1];
        else break;   //如果该点没有第2^(i - 1)个父亲了，也不会有更远的父亲
    }
    for(int i = head[u]; ~i; i = edge[i].next)  //遍历相邻的所有点
    {
        int v = edge[i].to;
        if(v != father) // v 不是 u 的父亲，就是 u 的儿子
        {
            dis[v] = dis[u] + edge[i].w;    //更新距离
            fa[v][0] = u;   // v 的第2^0个父亲即第一个父亲是 u 
            dfs(v, u);
        }
    }
}

int lca(int u, int v)
{
    if(dep[u] < dep[v]) //默认 u 比 v 深
        swap(u, v);
    for(int i = 20; i >= 0; --i)    //从大到小枚举使 x 和 y 到达同一层
    {
        if(dep[fa[u][i]] >= dep[v])
            u = fa[u][i];
        if(u == v)
            return u;
    }
    for(int i = 20; i >= 0; --i)
    {
        if(fa[u][i] != fa[v][i])
        {
            u = fa[u][i];
            v = fa[v][i];
        }
    }
    return fa[u][0];
}

int main()
{
    int t, u, v, w;
    scanf("%d", &t);
    while(t--)
    {
        init();
        scanf("%d%d", &n, &m);
        for(int i = 1; i < n; ++i)
        {
            scanf("%d%d%d", &u, &v, &w);
            add(u, v, w);
            add(v, u, w);
        }
        dfs(1, 0);  //选取1为根节点，1的父亲是0
        while(m--)
        {
            scanf("%d%d", &u, &v);
            cout<<dis[u] + dis[v] - 2 * dis[lca(u, v)]<<'\n';
        }
    }
    return 0;
}