NOIP2017 D2T1 奶酪

奶酪

题目背景：

NOIP2017 D2T1

分析：搜索

 

代码刚刚下来的时候，测试这道题，我差点还T了······不过事实证明还是没有卡正确复杂度的······比较简单，直接O(n²)判断一下，两个球是否联通，如果联通，则连边，然后在判断和上下底面相切的位置有哪些，将上下底面抽象成两个点，一样连边，然后从下底所代表的点，开始dfs，看是否能够到上底即可。

当时学军数据，最后两个点会爆掉long long，但是最后官方没有，感觉出题人的那个中间·····的确有逻辑漏洞······

 

Source：

/*
	created by scarlyw
*/
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <iostream>
#include <queue>
#include <set>
#include <vector>

template<class T>
inline void R(T &x) {
	static char c;
	static bool iosig;
	for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) {
		if (c == -1) return ;
		if (c == '-') iosig = true;
	}
	for (x = 0; isdigit(c); c = getchar())
		x = ((x << 2) + x << 1) + (c ^ '0');
	if (iosig) x = -x;
}

const int MAXN = 1000 + 10;

std::vector<int> edge[MAXN];
int n, h, r, t;
bool vis[MAXN];
struct ball {
	int x, y, z;
} b[MAXN];

inline void add_edge(int x, int y) {
	edge[x].push_back(y), edge[y].push_back(x);
}

inline bool able(int i, int j) {
	long long temp = (long long)(b[i].x - b[j].x) * (b[i].x - b[j].x)
				   + (long long)(b[i].y - b[j].y) * (b[i].y - b[j].y)
				   + (long long)(b[i].z - b[j].z) * (b[i].z - b[j].z);
	if ((long long)r * r * 4LL >= temp) return true;
	else return false;
}

inline void dfs(int cur) {
	vis[cur] = true;
	for (int p = 0; p < edge[cur].size(); ++p) {
		int v = edge[cur][p];
		if (!vis[v]) dfs(v);
	}
}

inline void solve() {
	R(n), R(h), R(r);
	memset(vis, 0, sizeof(bool) * (n + 5));
	for (int i = 0; i <= n + 1; ++i) edge[i].clear();
	for (int i = 1; i <= n; ++i) R(b[i].x), R(b[i].y), R(b[i].z);
	for (int i = 1; i <= n; ++i) if (b[i].z <= r) add_edge(0, i);
	for (int i = 1; i <= n; ++i) if (b[i].z >= h - r) add_edge(i, n + 1);
	for (int i = 1; i <= n; ++i)
		for (int j = i + 1; j <= n; ++j) 
			if (able(i, j)) add_edge(i, j);
	dfs(0);
	vis[n + 1] ? (std::cout << "Yes\n") : (std::cout << "No\n");
}

int main() {
	freopen("cheese.in", "r", stdin);
	freopen("cheese.out", "w", stdout);
	R(t);
	while (t--) solve();
	return 0;
}