[AcWing 3712] 根能抵达的点
二分 + BFS
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#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 20000 + 10;
int n, m;
int h[N], e[N], ne[N], w[N], idx;
bool st[N];
void add(int a, int b, int c)
{
e[idx] = b;
w[idx] = c;
ne[idx] = h[a];
h[a] = idx ++;
}
bool check(int x)
{
memset(st, false, sizeof st);
queue<int> q;
q.push(0);
int cnt = 1;
while (q.size()) {
auto t = q.front();
q.pop();
st[t] = true;
for (int i = h[t]; i != -1; i = ne[i]) {
int j = e[i];
if (st[j] || w[i] < x)
continue;
q.push(j);
cnt ++;
}
}
return cnt <= m;
}
void solve()
{
cin >> n >> m;
idx = 0;
memset(h, -1, sizeof h);
for (int i = 0; i < n - 1; i ++) {
int a, b, w;
cin >> a >> b >> w;
add(a, b, w);
}
int l = 0, r = 1e7;
while (l < r) {
int mid = l + r >> 1;
if (check(mid))
r = mid;
else
l = mid + 1;
}
cout << l << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T;
cin >> T;
while (T --) {
solve();
}
// solve();
return 0;
}
- \(check(mid)\) 用于检验当 \(X\) 设置为 \(mid\) 时,是否满足根节点能够抵达的节点数目(包括自己)不超过 \(Y\) 这个条件,如果满足,就让 \(r = mid\),在 \(check\) 函数内部,用 \(BFS\) 进行搜索,每次先把根节点放进队列,然后进行 \(BFS\),让边权大于等于 \(X\) 的边所连的点入队,统计一共有多少个点和根节点是连通的