HDU 4173 Party location (computational geometry, enumeration)

HDU 4173

Test instructions: Known n (n<=200) position of the contestant's residence coordinates, now invite as many players as possible to participate in a party, and each player for more than 2.5Km from the house is not to go to the party, for the maximum number of players can go to participate.

Ideas:

Consider the possible location of the party, and to invite as many players as possible, just consider where the party is located at the midpoint of a two-bit residence line or the location of a contestant's residence, as this is where the largest number of participants is likely to be located.

If the maximum number of participants can be obtained from other locations, then the midpoint or the contestant's residence will be available. Ok~, try to draw.

So, as long as we enumerate all the midpoint and the position of the contestant's residence, we can get the number of the party contestants, and the maximum value is the answer.

AC Code:

/** @author novicer* language:c++/c*/#include <iostream> #include <sstream> #include <fstream># include<vector> #include <list> #include <deque> #include <queue> #include <stack># include<map> #include <set> #include <bitset> #include <algorithm> #include <cstdio># include<cstdlib> #include <cstring> #include <cctype> #include <cmath> #include <ctime># Include<iomanip>using namespace Std;const double eps (1e-8); typedef long LONG lint;const int MAXN = +5;pair<do Uble,double>con[maxn];int Main () {//freopen ("Input.txt", "R", stdin), int n;while (CIN >> N) {for (int i = 1; I <= n; i++) scanf ("%lf%lf", &con[i].first,&con[i].second); int ans = 0;for (int i = 1; I <= n; i++) {//cout << con [I].first << "" << con[i].second << endl;for (int j = 1; J <= N; J + +) {pair<double,double> T;t.first = (Con[i].first + con[j].first)/2.0;t.second = (Con[i].second + con[J].second)/2.0;//cout << t.first << "" << t.second << endl;int cnt = 0;for (int k = 1; K <= N;  k++) {Double dis = POW (t.first-con[k].first,2) + POW (t.second-con[k].second,2);//cout << dis << endl;if (dis <= 6.25 + eps) cnt++;} ans = max (ans, CNT);}} cout << ans << endl;} return 0;}

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HDU 4173 Party location (computational geometry, enumeration)