题目描述 Description
小机房有棵焕狗种的树，树上有N个节点，节点标号为0到N-1，有两只虫子名叫飘狗和大吉狗，分居在两个不同的节点上。有一天，他们想爬到一个节点上去搞基，但是作为两只虫子，他们不想花费太多精力。已知从某个节点爬到其父亲节点要花费 c 的能量（从父亲节点爬到此节点也相同），他们想找出一条花费精力最短的路，以使得搞基的时候精力旺盛，他们找到你要你设计一个程序来找到这条路，要求你告诉他们最少需要花费多少精力
输入描述 Input Description
第一行一个n，接下来n-1行每一行有三个整数u, v, c 。表示节点 u 爬到节点 v 需要花费 c 的精力。
第n+1行有一个整数m表示有m次询问。接下来m行每一行有两个整数 u, v 表示两只虫子所在的节点
输出描述 Output Description
一共有m行，每一行一个整数，表示对于该次询问所得出的最短距离。
样例输入 Sample Input
3
1 0 1
2 0 1
3
1 0
2 0
1 2样例输出 Sample Output
1
1
2数据范围及提示 Data Size & Hint
1<=n<=50000，1<=m<=75000，0<=c<=1000
题解和代码
这道题显然是一道LCA的题目，我们可以跑一遍树上倍增。
#include <iostream>
#include <deque>
#include <vector>
#include <cmath>
using namespace std;
int n, m;
struct edge_type
{
int u, v, c, nextu, nextv;
};
edge_type edge[50005];
const int root = 0;
int hu[50005], hv[50005], fa[50005], fac[50005], deep[50005], f[50005][30], maxdeep = 0, fsum[50005][30];
int maxi;
bool vis[50005];
vector<int> son[50005];
deque<int> q;
void bfs_build_tree(const int root)
{
q.push_back(root);
for (int i = 0; i <= n; ++i)
{
vis[i] = false;
}
vis[root] = true;
deep[root] = 1;
int now;
while (!q.empty())
{
now = q.front();
if (maxdeep < deep[now])
maxdeep = deep[now];
q.pop_front();
int i = hu[now];
while (i)
{
if (!vis[edge[i].v])
{
q.push_back(edge[i].v);
vis[edge[i].v] = true;
fa[edge[i].v] = now;
fac[edge[i].v] = edge[i].c;
son[now].push_back(edge[i].v);
deep[edge[i].v] = deep[now] + 1;
}
i = edge[i].nextu;
}
i = hv[now];
while (i)
{
if (!vis[edge[i].u])
{
q.push_back(edge[i].u);
vis[edge[i].u] = true;
fa[edge[i].u] = now;
fac[edge[i].u] = edge[i].c;
son[now].push_back(edge[i].u);
deep[edge[i].u] = deep[now] + 1;
}
i = edge[i].nextv;
}
}
}
void get_f()
{
maxi = log2(double(maxdeep)-0.5);
for (int j = 0; j < n; ++j)
{
f[j][0] = fa[j];
fsum[j][0] = fac[j];
for (int i = 1; i <= 30; ++i)
fsum[j][i] = f[j][i] = -1;
}
fsum[0][0] = f[0][0] = -1;
for (int i = 1; i <= maxi; ++i)
for (int j = 1; j < n; ++j)
if ((1 << i) < deep[j])
{
f[j][i] = f[f[j][i-1]][i-1];
fsum[j][i] = fsum[j][i-1] + fsum[f[j][i-1]][i-1];
}
}
int g(int x, int y)
{
int ret = 0, a = x, b = y;
while (deep[b] > deep[a])
{
for (int i = maxi; i >= 0; --i)
if (deep[f[b][i]] > deep[a])
{
ret += fsum[b][i];
b = f[b][i];
}
if (deep[fa[b]] == deep[a])
{
ret += fac[b];
b = fa[b];
}
}
while (deep[a] > deep[b])
{
for (int i = maxi; i >= 0; --i)
if (deep[f[a][i]] > deep[b])
{
ret += fsum[a][i];
a = f[a][i];
}
if (deep[fa[a]] == deep[b])
{
ret += fac[a];
a = fa[a];
}
}
while (a != b)
{
for (int i = maxi; i >= 0; --i)
if (f[a][i] != f[b][i])
{
ret += (fsum[a][i] + fsum[b][i]);
a = f[a][i];
b = f[b][i];
}
if (fa[a] == fa[b])
{
ret += (fac[a] + fac[b]);
a = fa[a];
b = fa[b];
}
}
return ret;
}
int main()
{
cin >> n;
int u, v, c;
for (int i = 1; i < n; ++i)
{
cin >> u >> v >> c;
edge[i].u = u;
edge[i].v = v;
edge[i].c = c;
edge[i].nextu = hu[u];
edge[i].nextv = hv[v];
hu[u] = i;
hv[v] = i;
}
bfs_build_tree(root);
get_f();
cin >> m;
int a, b;
for (int i = 0; i < m; ++i)
{
cin >> a >> b;
cout << g(a, b) << endl;
}
}