풀이
노가다를 해보자~
사실 노가다 안 해도 됨
코드
#include <iostream>
#include <string>
using namespace std;
int x[5], y[5];
string gogo() {
if (x[0] == x[3] && y[0] == y[3] ||
x[1] == x[2] && y[0] == y[3] ||
x[0] == x[3] && y[1] == y[2] ||
x[1] == x[2] && y[1] == y[2]) return "POINT";
if (x[0] == x[3] && y[2] <= y[0] && y[0] <= y[3] ||
x[0] == x[3] && y[2] <= y[1] && y[1] <= y[3] ||
x[1] == x[2] && y[2] <= y[0] && y[0] <= y[3] ||
x[1] == x[2] && y[2] <= y[1] && y[1] <= y[3] ||
x[0] == x[3] && y[0] <= y[2] && y[2] <= y[1] ||
x[0] == x[3] && y[0] <= y[3] && y[3] <= y[1] ||
x[1] == x[2] && y[0] <= y[2] && y[2] <= y[1] ||
x[1] == x[2] && y[0] <= y[3] && y[3] <= y[1] ||
y[0] == y[3] && x[2] <= x[0] && x[0] <= x[3] ||
y[0] == y[3] && x[2] <= x[1] && x[1] <= x[3] ||
y[1] == y[2] && x[2] <= x[0] && x[0] <= x[3] ||
y[1] == y[2] && x[2] <= x[1] && x[1] <= x[3] ||
y[0] == y[3] && x[0] <= x[2] && x[2] <= x[1] ||
y[0] == y[3] && x[0] <= x[3] && x[3] <= x[1] ||
y[1] == y[2] && x[0] <= x[2] && x[2] <= x[1] ||
y[1] == y[2] && x[0] <= x[3] && x[3] <= x[1]) return "LINE";
if (x[2] <= x[0] && x[0] <= x[3] && y[2] <= y[0] && y[0] <= y[3] ||
x[2] <= x[1] && x[1] <= x[3] && y[2] <= y[1] && y[1] <= y[3] ||
x[2] <= x[0] && x[0] <= x[3] && y[2] <= y[1] && y[1] <= y[3] ||
x[2] <= x[1] && x[1] <= x[3] && y[2] <= y[0] && y[0] <= y[3] ||
x[0] <= x[2] && x[2] <= x[1] && y[0] <= y[2] && y[2] <= y[1] ||
x[0] <= x[3] && x[3] <= x[1] && y[0] <= y[3] && y[3] <= y[1] ||
x[0] <= x[2] && x[2] <= x[1] && y[0] <= y[3] && y[3] <= y[1] ||
x[0] <= x[3] && x[3] <= x[1] && y[0] <= y[2] && y[2] <= y[1] ||
x[0] <= x[2] && x[3] <= x[1] && y[2] <= y[0] && y[1] <= y[3] ||
x[0] <= x[2] && x[3] <= x[1] && y[2] <= y[1] && y[1] <= y[3] ||
x[0] <= x[2] && x[3] <= x[1] && y[2] <= y[0] && y[0] <= y[3] ||
x[2] <= x[0] && x[1] <= x[3] && y[0] <= y[2] && y[3] <= y[1] ||
x[2] <= x[1] && x[1] <= x[3] && y[0] <= y[2] && y[3] <= y[1] ||
x[2] <= x[0] && x[0] <= x[3] && y[0] <= y[2] && y[3] <= y[1]) return "FACE";
return "NULL";
}
int main() {
for (int i = 0; i < 4; i++) cin >> x[i] >> y[i];
cout << gogo();
return 0;
}
아무말
백준, 백준 온라인 저지, BOJ, Baekjoon Online Judge, C, C++, 씨, 씨쁠쁠, JAVA, algorithm, 자바, 알고리즘, 자료구조, 문제, 문제 풀이, 풀이