每次寻找最大深度的节点,若未被覆盖则将其爷爷设为站点并更新父辈的距离。
其中(d[i])是该点的深度,(dis[i])是它到最近的消防站的距离。
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 1e3 + 5;
int n, f[maxn], ans;
int d[maxn], b[maxn], dis[maxn];
int main() {
scanf("%d", &n);
for (int i = 2; i <= n; i++) {
scanf("%d", &f[i]);
d[i] = d[f[i]] + 1;
}
for (int i = 0; i <= n; i++) {
b[i] = i;
dis[i] = maxn;
}
sort(b + 1, b + 1 + n, [](int x, int y){ return d[x] > d[y]; });
for (int i = 1; i <= n; i++) {
int cur = b[i], fa = f[cur], gf = f[fa];
dis[cur] = min(dis[cur], min(dis[fa] + 1, dis[gf] + 2));
if (dis[cur] > 2) {
ans++;
dis[gf] = 0;
dis[f[gf]] = min(dis[f[gf]], 1);
dis[f[f[gf]]] = min(dis[f[f[gf]]], 2);
}
}
return !printf("%d
", ans);
}