hdu2586 How far away ？（tarjin求LCA）

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

ECJTU 2009 Spring Contest

 

題意：給出房屋的個數 n ，邊數爲 n - 1，詢問 m 次，每次詢問兩房屋之間的路徑長度

 

思路：數據比較大，用 tarjin 離線求 lca，過程中維護距離值，dis[i]爲點 i 到根節點1的距離

 

tarjin求lca講解：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;
}