UVa 511

这题不难,但细节比较多,排序的依据有很多,比较容易出错。考差C++和STL也比较全面,比如结构体、构造函数、map、pair、vector、sort、unique、copy等等,使用C++11的lambda配合泛型算法是一个很好的选择。

#include <bits/stdc++.h>
#define EPS 1e-7
using namespace std;

struct MAP
{
	string name;
	double x1, y1, x2, y2, area, Ratio;
	pair<double, double> center;
	int ilevel = -1;
	MAP(double _x1, double _y1, double _x2, double _y2, string _name) :
		x1(_x1), y1(_y1), x2(_x2), y2(_y2), name(_name) {
		if (x1 > x2) swap(x1, x2);
		if (y1 > y2) swap(y1, y2);
		area = (x2 - x1) * (y2 - y1);
		center.first = (x1 + x2) / 2.0;
		center.second = (y1 + y2) / 2.0;
		Ratio = (y2 - y1) / (x2 - x1);
	}
	bool contain(pair<double, double> x) const {
		return x.first >= x1 && x.first <= x2 && x.second >= y1 && x.second <= y2;
	}
};
vector<MAP> maps;
map<string, pair<double, double>> site;
pair<double, double> Qpos;

double dist(pair<double, double> a, pair<double, double> b) {
	return hypot(a.first - b.first, a.second - b.second);
}
bool cmp(const MAP & a, const MAP & b){
	if (a.ilevel != b.ilevel) return a.ilevel > b.ilevel;
	double d1 = dist(a.center, Qpos), d2 = dist(b.center, Qpos);
	if (fabs(d1 - d2) > EPS) return d1 < d2;
	d1 = fabs(a.Ratio - 0.75), d2 = fabs(b.Ratio - 0.75);
	if (fabs(d1 - d2) > EPS) return d1 < d2;
	d1 = dist(Qpos, make_pair(a.x2, a.y1)), d2 = dist(Qpos, make_pair(a.x2, a.y1));
	if (fabs(d1 - d2) > EPS) return d1 > d2;
	return a.x1 < b.x1;
}
int main()
{
	ios::sync_with_stdio(false);
	string name; cin >> name;
	while (cin >> name, name != "LOCATIONS") {
		double x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
		maps.push_back(MAP(x1, y1, x2, y2, name));
	}
	while (cin >> name, name != "REQUESTS") {
		double x, y; cin >> x >> y;
		site[name] = make_pair(x, y);
	}
	while (cin >> name, name != "END") {
		int level; cin >> level;
		if (!site.count(name)) printf("%s at detail level %d unknown location
", name.c_str(), level);
		else {
			printf("%s at detail level %d ", name.c_str(), level);
			Qpos = site[name];
			vector<double> all_area;
			vector<MAP> C1, C2;
			for (auto & i : maps) if (i.contain(Qpos)) {
				C1.push_back(i);
				all_area.push_back(i.area);
			}
			sort(all_area.begin(), all_area.end(), greater<double>());
			unique(all_area.begin(), all_area.end(), greater<double>());
			for (auto & i : C1)
				i.ilevel = find_if(all_area.begin(), all_area.end(), [i](const double area){return fabs(area - i.area) <= EPS; }) - all_area.begin() + 1;
			sort(C1.begin(), C1.end(), cmp);
			copy_if(C1.begin(), C1.end(), back_inserter(C2), [level](const MAP & x) {return x.ilevel == level; });
			if (C1.size() == 0) puts("no map contains that location");
			else if (C2.size() == 0) printf("no map at that detail level; using %s
", C1[0].name.c_str());
			else printf("using %s
", C2[0].name.c_str());
		}
	}
	return 0;
}


原文地址:https://www.cnblogs.com/kunsoft/p/5312704.html