hdu 2586 How far away ？(LCA)

hdu 2586 How far away ？(LCA)

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How far away ？

Time Limit: 2000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 559    Accepted Submission(s): 218

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

 

 

题目大意：
n个点，n-1条路形成了一棵树，然后又m个询问：(x,y)输出x到y的最短距离，典型的LCA问题，用Tarjan解决，时间复杂度为O(n+m)
代码：

#include  < iostream >
#include  < stdio.h >
using   namespace  std;

#define  max_size 40002

int  cnt, head[max_size], next[max_size];
int   set [max_size], d[max_size], ans[ 201 ];
bool  fals[max_size];

struct  Edge
{
    int v;
    int w;
    int pre;
} eg[ 2 * max_size], ask[ 404 ];

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


void  add_edge( int  x,  int  y,  int  z)
{
    eg[cnt].v = y;
    eg[cnt].w = z;
    eg[cnt].pre = head[x];
    head[x] = cnt++;
}

void  add_ask( int  x,  int  y,  int  cast)
{
    ask[cnt].v = y;
    ask[cnt].w = cast;
    ask[cnt].pre = next[x];
    next[x] = cnt++;
}

void  Tarjan( int  k)
{
    fals[k] = true;

    if (head[k] == 0)return ;

    int i, x, root;

    for (i = next[k]; i != 0; i = ask[i].pre)
    {
        x = ask[i].v;
        if (!fals[x])continue;
        root = find(x);
        ans[ask[i].w] = d[x] + d[k] - 2*d[root];
    }

    for (i = head[k]; i != 0; i = eg[i].pre)
    {
        x = eg[i].v;
        if (fals[x])continue;
        d[x] = d[k] + eg[i].w;
        Tarjan(x);
        x = find(x);
        set[x] = k;
    }

}

int  main()
{
    int i, m, n;
    int x, y, z;
    int t;
    cin >> t;
    while (t--)
    {
        cin >> n >> m;
        for (i = 0; i <= n; i++)
        {
            head[i] = next[i] = 0;
            set[i] = i;
            fals[i] = false;
        }
        
        cnt = 1;
        for (i = 1; i < n; i++)
        {
            scanf("%d%d%d", &x, &y, &z);
            add_edge(x, y, z);
            add_edge(y, x, z);
        }
        cnt = 1;
        for (i = 1; i <= m; i++)
        {
            scanf("%d%d", &x, &y);
            add_ask(x, y, i);
            add_ask(y, x, i);
        }
        d[1] = 0;
        Tarjan(1);

        for (i = 1; i <= m; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}