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#include <bits/stdc++.h> | |
using namespace std; | |
const int MAXN = 132694; | |
int n; | |
struct retangulo{ // criamos uma struct para representar os retangulos | |
long long int x1, x2, y1, y2; | |
bool operator == (retangulo r) const{ | |
if(x1 == r.x1 && x2 == r.x2 && y1 == r.y1 && y2 == r.y2) return true; | |
return false; | |
} | |
}; | |
retangulo v[MAXN]; // o vetor v guarda os retangulos originais | |
retangulo pref[MAXN], suf[MAXN]; | |
retangulo nullrec; | |
retangulo interseccao(retangulo r1, retangulo r2){ | |
//verificamos se a intersecção existe | |
if(r1 == nullrec || r2 == nullrec) return nullrec; | |
if(r1.x1 > r2.x2 || r2.x1 > r1.x2) return nullrec; | |
if(r1.y1 > r2.y2 || r2.y1 > r1.y2) return nullrec; | |
//guardamos as coordenadas da interseccao | |
int x1 = max(r1.x1, r2.x1); | |
int x2 = min(r1.x2, r2.x2); | |
int y1 = max(r1.y1, r2.y1); | |
int y2 = min(r1.y2, r2.y2); | |
//retornamos a interseccao | |
retangulo resp; | |
resp.x1 = x1; | |
resp.x2 = x2; | |
resp.y1 = y1; | |
resp.y2 = y2; | |
return resp; | |
} | |
int main() { | |
//deixamos o retangulo nulo com dimensoes inalcancaveis pelas restricoes | |
nullrec.x1 = 11234567890; | |
nullrec.x2 = 11234567890; | |
nullrec.y1 = 11234567890; | |
nullrec.y2 = 11234567890; | |
scanf("%d", &n); | |
for(int i = 0; i < n; i++ ){ | |
scanf("%lld %lld %lld %lld", &v[i].x1, &v[i].y1, &v[i].x2, &v[i].y2); | |
} | |
pref[0] = v[0]; | |
//calculamos o prefixo dos retangulos dados | |
for(int i = 1; i < n; i++) | |
pref[i] = interseccao(pref[i-1], v[i]); | |
if(!(pref[n-2] == nullrec)){ | |
printf("%lld %lld\n", pref[n-2].x1, pref[n-2].y1); | |
return 0; | |
} | |
suf[n - 1] = v[n - 1]; | |
//calculamos o sufixo dos retangulos dados | |
for(int i = n-2; i >= 0; i--) | |
suf[i] = interseccao(suf[i+1], v[i]); | |
if(!(suf[1] == nullrec)){ | |
printf("%lld %lld\n", suf[1].x1, suf[1].y1); | |
return 0; | |
} | |
// se chegamos aqui, temos que iterar pelo sufixo e prefixo para achar a resposta | |
for(int i = 1; i < n-1; i++){ | |
retangulo r = interseccao(pref[i-1], suf[i+1]); | |
if(!(r == nullrec) ){ | |
printf("%lld %lld\n", r.x1, r.y1); | |
break; | |
} | |
} | |
return 0; | |
} |
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