HDU 1875 畅通工程再续

最新推荐文章于 2021-07-27 08:46:09 发布

原创最新推荐文章于 2021-07-27 08:46:09 发布 · 176 阅读

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本文探讨了Kruskal算法在解决特定条件下的建桥问题上的应用，通过构建并查找连通组件来确定最小生成树，确保岛屿间桥梁建设的最优化。使用C++实现算法，并详细解释了代码逻辑。

一、题目描述

在这里插入图片描述

二、算法分析说明与代码编写指导

只对距离在 [10, 1000] 范围内的岛之间建桥。建完以后跑一次 Kruskal，判断是否有生成树（选中的边数是否达到 C - 1）。

三、AC 代码

#include<cstdio>
#include<algorithm>
#include<cmath>
#pragma warning(disable:4996)
using namespace std;
template<size_t n> class union_find {
private:
	unsigned root[n]; int rank[n];
public:
	union_find<n>() { init(); }
	union_find<n>(const bool& WannaInit) { if (WannaInit == true)init(); }
	void init() {
		fill(rank, rank + n, 1); for (unsigned i = 0; i < n; ++i)root[i] = i;
	}
	void init(const size_t& _n) {
		fill(rank, rank + _n, 1); for (unsigned i = 0; i < _n; ++i)root[i] = i;
	}
	unsigned find_root(const unsigned& v) {
		unsigned r = v, t = v, u;
		if (t == root[v])return v;
		while (r != root[r]) { r = root[r]; }
		while (t != r) { u = root[t]; root[t] = r; t = u; }
		return r;
	}
	void path_compress() const { for (unsigned i = 0; i < n; ++i)find_root(i); }
	void merge(unsigned u, unsigned v) {
		unsigned fu = find_root(u), fv = find_root(v); int d = rank[fu] - rank[fv];
		if (d < 0) { swap(fu, fv); swap(u, v); }
		else if (d == 0)++rank[fu];
		root[fv] = fu;
	}
	void merge_no_path_compression(const unsigned& u, const unsigned& v) {
		root[v] = u; if (rank[u] == rank[v])++rank[u];
	}
	void merge_directly(const unsigned& u, const unsigned& v) { root[v] = u; }
	unsigned _rank(const unsigned& v) const { return rank[find_root(v)]; }
	size_t size() const { return n; }
};
struct point { int x, y; }; struct edge { unsigned u, v; double w; };
inline bool cmp(const edge& e, const edge& f) { return e.w < f.w; }
inline double dist(const point& p, const point& q) { return hypot(p.x - q.x, p.y - q.y); }
unsigned t, c, n; point p[100], * P; edge e[4950], * E; double d, a; union_find<100> u;
int main() {
	scanf("%u", &t); ++t;
	while (--t) {
		scanf("%u", &c); E = e;
		for (unsigned i = 0; i < c; ++i) {
			P = p + i; scanf("%u%u", &P->x, &P->y);
			for (unsigned j = 0; j < i; ++j) {
				if ((d = dist(*P, p[j])) >= 10 && d <= 1000) {
					E->u = i; E->v = j; E->w = d * 100; ++E;
				}
			}
		}
		u.init(c); sort(e, E, cmp); n = 0; --c; a = 0;
		for (auto i = e; n < c && i != E; ++i) {
			if (u.find_root(i->u) != u.find_root(i->v)) { u.merge(i->u, i->v); ++n; a += i->w; }
		}
		if (n == c)printf("%.1lf\n", a);
		else puts("oh!");
	}
	return 0;
}