Solução por João Guilherme.
Para ver o problema original, clique aqui.
Esse problema é muito semelhante ao resolvido na semana 21, para ver a solução clique aqui. Para resolver este problema só precisamos guardar o tipo da figura num vetor e checar se o ponto está na figura de acordo com o seu tipo.
Segue o código para melhor entendimento.
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| #include <bits/stdc++.h> | |
| #define ff first | |
| #define ss second | |
| using namespace std; | |
| typedef pair <double, double> pdd; | |
| typedef pair <pdd, pdd> ppd; | |
| typedef pair <ppd, char> fig; | |
| fig arr[16]; | |
| double dist(pdd a, pdd b){ | |
| return sqrt((a.ff - b.ff)*(a.ff - b.ff) + (a.ss - b.ss)*(a.ss - b.ss)); | |
| } | |
| int main(){ | |
| ios_base::sync_with_stdio(0); | |
| int t; | |
| char op; | |
| bool vdc; | |
| pdd points; | |
| cin >> op; | |
| for(t = 1; op != '*'; ++t){ | |
| arr[t].ss = op; | |
| if(op == 'r') | |
| cin >> arr[t].ff.ff.ff >> arr[t].ff.ff.ss >> arr[t].ff.ss.ff >> arr[t].ff.ss.ss; | |
| else | |
| cin >> arr[t].ff.ff.ff >> arr[t].ff.ff.ss >> arr[t].ff.ss.ff; | |
| cin >> op; | |
| } | |
| cin >> points.ff >> points.ss; | |
| for(int i = 1; points.ff != 9999.9 || points.ss != 9999.9; ++i){ | |
| vdc = 0; | |
| for(int j = 1; j < t; ++j){ | |
| if(arr[j].ss == 'r'){ | |
| if(points.ff > arr[j].ff.ff.ff && points.ss < arr[j].ff.ff.ss && points.ff < arr[j].ff.ss.ff && points.ss > arr[j].ff.ss.ss){ | |
| vdc = 1; | |
| cout << "Point "<< i <<" is contained in figure " << j << "\n"; | |
| } | |
| } | |
| else | |
| if(dist(points, arr[j].ff.ff) < arr[j].ff.ss.ff){ | |
| vdc = 1; | |
| cout << "Point "<< i <<" is contained in figure " << j << "\n"; | |
| } | |
| } | |
| if(!vdc) cout << "Point " << i <<" is not contained in any figure\n"; | |
| cin >> points.ff >> points.ss; | |
| } | |
| return 0; | |
| } |