【37.48%】【hdu 2587】How far away ？(3篇文章,3种做法,LCA之树上倍增)

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
",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
", dis[pret0]+dis[pret1]-2*dis[p[t0][0]]);
        }
    }
    return 0;
}