【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\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;
}