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

Total Submission(s): 13390 Accepted Submission(s): 5018

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(0k<=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

【题解】



设p[i][j]表示i往上走2^j个节点到达的节点。

p[i][j]=p[p[i][j-1]][j-1];

然后先让所求的两个点的高度一样。

即高度高的一直往上走走到和高度底的高度一样;

然后再一起往上走走到共同的祖先；(最近);

找到祖先后输出dis[x]+dis[y]-2*dis[LCA];就是距离了。

这个距离是树上的最短距离。还是很有用的。可以扩展下；

#include <cstdio>

#include <vector>

#include <iostream>

#include <algorithm>

using namespace std;

const int MAXN = 50000;

const int MAX = 16;

vector <int> son[MAXN],w[MAXN];

int n,p[MAXN][MAX+5],dep[MAXN],pre[MAX+5],m;

long long dis[MAXN];

void input(int &r)

{

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    r = 0;

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

}

void dfs(int x,int f)

{

    dep[x] = dep[f] + 1;

    p[x][0] = f;

    for (int i = 1; i <= MAX; i++)

        p[x][i] = p[p[x][i - 1]][i - 1];

    int len = son[x].size();

    for (int i = 0; i <= len - 1; i++)

    {

        int y = son[x][i];

        if (y != f)

        {

            dis[y] = dis[x] + w[x][i];

            dfs(y, x);

        }

    }

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    pre[0] = 1;

    for (int i = 1; i <= MAX; i++)

        pre[i] = pre[i - 1] << 1;

    int T;

    input(T);

    while (T--)

    {

        input(n); input(m);

        for (int i = 1; i <= n; i++)

            son[i].clear(),w[i].clear();

        for (int i = 1; i <= n - 1; i++)

        {

            int x, y, z;

            input(x); input(y); input(z);

            son[x].push_back(y);

            w[x].push_back(z);

            son[y].push_back(x);

            w[y].push_back(z);

        }

        dis[1] = 0;

        dfs(1, 0);

        for (int i = 1; i <= m; i++)

        {

            int t0, t1,pret0,pret1;

            input(t0); input(t1);

            pret0 = t0; pret1 = t1;

            if (dep[t0] > dep[t1])

                swap(t0, t1);

            for (int i = MAX; i >= 0; i--)

                if (dep[t0] <= dep[t1] - pre[i])

                    t1 = p[t1][i];

            if (t1 == t0)

            {

                printf("%I64d\n",dis[pret0]+dis[pret1]-2*dis[t0]);

                continue;

            }

            for (int i = MAX; i >= 0; i--)

            {

                if (p[t0][i] == p[t1][i])

                    continue;

                t0 = p[t0][i], t1 = p[t1][i];

            }

            printf("%I64d\n", dis[pret0]+dis[pret1]-2*dis[p[t0][0]]);

        }

    }

    return 0;

}