Algorithm
Problem Name: 2 AD-HOC - beecrowd | 1618
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/1618
Colision
By Jony Teixeira de Melo, Centro Universitário do Triângulo 
Brazil
Timelimit: 1
You have been tasked to check if the robot invaded a rectangular area formed by four cardinal points (A, B, C and D). Will be informed the four cardinal points of a plan as shown. The area is formed by connecting the four points as follows AB, BC, CD and DA. The X, Y coordinate of the robot will be informed
Input
The input consists of several test cases. The first line is formed by a number N indicating the total number of test cases. The next N lines consist of 10 integers (Ax, Ay, Bx, By, Cx, Cy, Dx, Dy, RX, RY) representing each of the vertices A, B, C and D and the location X, Y robot. Each value is separated by a blank space.
Output
The output should print for each test case, the number 1 if the robot is within the area (considering the edges of the figure as part of the area of the figure) , and print the number 0 otherwise.
| Sample Input | Sample Output |
|
5 3 6 6 6 6 5 3 5 5 4 1 1 7 1 7 7 1 7 4 2 1 4 7 4 7 6 1 6 5 5 6 2 9 2 9 6 6 6 1 7 4 3 9 3 9 5 4 5 10 7 |
0 1 1 0 0 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main(void) {
int n, i;
int ax, ay, bx, by, cx, cy, dx, dy, rx, ry;
scanf("%d", &n);
for (i = 0; i < n; ++i) {
scanf("%d %d %d %d %d", &ax, &ay, &bx, &by, &cx);
scanf("%d %d %d %d %d", &cy, &dx, &dy, &rx, &ry);
if (dy >= ry && cy >= ry && ay < = ry && by <= ry)
if (dx <= rx && ax <= rx && cx >= rx && bx >= rx)
printf("1");
else printf("0");
else printf("0");
printf("\n");
}
return 0;
}
Input
3 6 6 6 6 5 3 5 5 4
1 1 7 1 7 7 1 7 4 2
1 4 7 4 7 6 1 6 5 5
6 2 9 2 9 6 6 6 1 7
4 3 9 3 9 5 4 5 10 7
Output
1
1
0
0
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <iostream>
using namespace std;
int main(void) {
int n, i;
int ax, ay, bx, by, cx, cy, dx, dy, rx, ry;
cin >> n;
for (i = 0; i < n; ++i) {
cin >> ax >> ay >> bx >> by >> cx >> cy >> dx >> dy >> rx >> ry;
if (dy >= ry and cy >= ry and ay < = ry and by <= ry)
if (dx <= rx and ax <= rx and cx >= rx and bx >= rx)
cout << 1;
else cout << 0;
else cout << 0;
cout << endl;
}
return 0;
}
Input
3 6 6 6 6 5 3 5 5 4
1 1 7 1 7 7 1 7 4 2
1 4 7 4 7 6 1 6 5 5
6 2 9 2 9 6 6 6 1 7
4 3 9 3 9 5 4 5 10 7
Output
1
1
0
0