How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6422    Accepted Submission(s): 2411

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2586

题目大意：给一个带权树，m次询问，求两点之间的距离

题目分析：裸的LCA问题，采用离线的Tarjan算法

#include <cstdio>
#include <cstring>
int const MAX = 40005;

struct Edge
{
    int id, val; //当前边序号，边权
    int next;  //下一条
}e[2 * MAX];

int n, m, cnt;
//x, y表示询问的起点和终点，z是x和y的LCA
int x[MAX], y[MAX], z[MAX];
//fa存祖先，dist存到根的距离，pre存父亲
int fa[MAX], dist[MAX], pre[MAX];
bool vis[MAX];

void AddEdge(int u, int v, int w)
{
    e[cnt].id = u;
    e[cnt].val = w;
    e[cnt].next = pre[v];
    pre[v] = cnt++;

    e[cnt].id = v;
    e[cnt].val = w;
    e[cnt].next = pre[u];
    pre[u] = cnt++;
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void tarjan(int k)
{
    vis[k] = true;
    fa[k] = k;
    for(int i = 1; i <= m; i++)
    {
        if(x[i] == k && vis[y[i]])
            z[i] = Find(y[i]);
        if(y[i] == k && vis[x[i]])
            z[i] = Find(x[i]);
    }
    for(int i = pre[k]; i != -1; i = e[i].next)
    {
        if(!vis[e[i].id])
        {
            dist[e[i].id] = dist[k] + e[i].val;
            tarjan(e[i].id);
            fa[e[i].id] = k;
        }
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int u, v, w;
        scanf("%d %d", &n, &m);
        cnt = 0;
        memset(pre, -1, sizeof(pre));
        for(int i = 1; i < n; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            AddEdge(u, v, w);
        }
        for(int i = 1; i <= n; i++)
            x[i] = y[i] = z[i] = 0;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d %d", &u, &v);
            x[i] = u;
            y[i] = v;
        }
        memset(vis, false, sizeof(vis));
        dist[1] = 0;
        tarjan(1);
        for(int i = 1; i <= m; i++)
            printf("%d\n",dist[x[i]] + dist[y[i]] - 2 * dist[z[i]]);
    }
}