[Luogu] P3958 奶酪

Luogu P3958 奶酪

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传送门

这个题暴力可过【手动微笑】。好像据我知道的有BFS啦,DFS啦，还有并查集。不管你怎么写，反正我写的DFS【再次手动微笑】。首先读入数据，遍历每一个未访问的空洞进行DFS，再按距离判断哪个空洞可以继续到达，最后判断圆心高度加半径是否大于等于奶酪高度即可(即\(hole[i].z + r \geqslant h\))。哎，可怜一个水题，我竟然考场上爆炸，还是太水啦。

#include <algorithm>
#include <cstdio>
#include <cmath>
typedef long long ll;
int T, n, h, r, stp, vis[1001];
bool fnd;
double tmp;
struct che{
	int x, y, z;
	bool operator < (const che &chi)const{return z < chi.z;}
}c[1001];
inline double dist(int now, int i) {
	return sqrt(double(c[now].x - c[i].x) * (c[now].x - c[i].x) + 
	double(c[now].y - c[i].y) * (c[now].y - c[i].y) +
	double(c[now].z - c[i].z) * (c[now].z - c[i].z));
}
void dfs(int now) {
	if(c[now].z + r >= h) {
		fnd = true;
		return ;
	}
	vis[now] = stp;
	for (int i = 1; i <= n; ++i) {
		tmp = dist(now, i);
		if(tmp <= r * 2 && vis[i] != stp && !fnd)
			dfs(i);
	}
	return ;
}
void solve() {
	fnd = false;
	scanf("%d%d%d", &n, &h, &r);
	for (int i = 1; i <= n; ++i)
		scanf("%d%d%d", &c[i].x, &c[i].y, &c[i].z);
	std::sort(c + 1, c + n + 1);
	for (int i = 1; i <= n; ++i) 
		if(!fnd && std::abs(c[i].z) - r <= 0 && vis[i] != stp)
			dfs(i);
	if(fnd) printf("Yes\n");
	else printf("No\n");
	return ;
}
int main() {
	scanf("%d", &T);
	while(T--) {
		stp++;
		solve();
	}
	return 0;
}