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  • 洛谷2279(贪心)

    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);
    }
    
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  • 原文地址:https://www.cnblogs.com/AlphaWA/p/10856579.html
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