#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
typedef pair<int, int> PII;
const int N = 10010, M = N * 2;
int n, m;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
int p[N];
int res[M * 2];
int st[N];
vector<PII> query[N]; // first存查询的另外一个点,second存查询编号
void add(int a, int b, int c)
{
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx ++ ;
}
void dfs(int u, int fa)
{
for (int i = h[u]; ~i; i = ne[i])
{
int j = e[i];
if (j == fa) continue;
dist[j] = dist[u] + w[i];
dfs(j, u);
}
}
int find(int x)
{
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
void tarjan(int u)
{
st[u] = 1;
for (int i = h[u]; ~i; i = ne[i])
{
int j = e[i];
if (!st[j])
{
tarjan(j);
p[j] = u;
}
}
for (auto item : query[u])
{
int y = item.first, id = item.second;
if (st[y] == 2)
{
int anc = find(y);
res[id] = dist[u] + dist[y] - dist[anc] * 2;
}
}
st[u] = 2;
}
int main()
{
scanf("%d%d", &n, &m);
memset(h, -1, sizeof h);
for (int i = 0; i < n - 1; i ++ )
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c), add(b, a, c);
}
for (int i = 0; i < m; i ++ )
{
int a, b;
scanf("%d%d", &a, &b);
if (a != b)
{
query[a].push_back({b, i});
query[b].push_back({a, i});
}
}
for (int i = 1; i <= n; i ++ ) p[i] = i;
dfs(1, -1);
tarjan(1);
for (int i = 0; i < m; i ++ ) printf("%d\n", res[i]);
return 0;
}
tarjan();
为什么放在p[j] = u;
的前面啊?
同有疑惑
这是我个人的看法:如果交换顺序,2号点将无法被并查集合并,必须回溯完再合并,就是说需要把当前节点的所有子节点都处理完,再都合并到父节点中。如果顺序反了,每个节点在并查集的根节点都是它的父节点,都属于独立的集合,就无法正确计算最近公共祖先了,除非这棵树只有一层。