维护一个dist数组,dist[i]表示节点 i 到黑色节点的最短距离(如果 i 已经是黑色节点,为0)
从ci开始搜索,把ci点染黑后,bfs更新搜索到的点的dist,并入队
用下一次染黑的节点dist更新ans,并输出答案
时间复杂度O(NN)
代码
#include<bits/stdc++.h>
using namespace std;#define CaseT int CaseT; cin >> CaseT; while(CaseT--)
const int N = 2e5 + 10;
int c[N];
int dist[N];
void solve() {
int n; cin >> n;
vector<vector<int>> g(n + 1);
for (int i = 1; i <= n; i++) {
cin >> c[i];
}
int m = n - 1;
while (m--) {
int u, v;
cin >> u >> v;
g[u].push_back(v), g[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
dist[i] = n;
}
int ans = n;
for (int i = 1; i <= n; i++) {ans = min(ans, dist[c[i]]);
if (i > 1) {
ans = min(ans, dist[c[i]]);
cout << ans << " ";
}
queue<int> q;
q.push(c[i]);
dist[c[i]] = 0;
while (q.size()) {
int u = q.front(); q.pop();
for (auto v : g[u]) {
if (dist[u] + 1 < min(dist[v], ans)) {
dist[v] = dist[u] + 1;
q.push(v);
}
}
}
}
cout << endl;
}
int main() {
CaseT
solve();
}