Different divisiontime limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 234 accepted submission (s): 90
Problem descriptionnow we will give you a graph, there are getting points in the graph. we will choose two different points arbitrarily, and connect them as a line. please tell us that whether these points (include the two points referred above) is on the left side of the line, or lying on the line. or on the right side of the line. for example,
There are four points in the graph: A, B, C, D. we connect C and D. now C and D form a new line "cd ". obviusly, C and D are lying on the line "CD ". A is on the right side of Cd, and B is on the left side of CD. what's more, A is on the left side of line DC, and B is on the right side of line DC. so line "CD" and "DC" are different in this problem;
Inputthe first line of input is a single integer T, indicating the number of test cases. then exactly t test cases followed. in each case, the first line contains one INTEGER: N, the number of points. then n lines followed, each line contains two real numbers x, y, indicating the coordinates of points. then one line follows, contains two integers P1 and P2 indicate the p1th point and the p2th point in this case.
1 <= T <= 100
2 <= n <= 1000
1 <= p1, p2 <= N, P1! = P2
-1000 <X, Y <1000;
Outputfor each case, print n lines. according to the order of input, for each point print "left" if this point is on the left side of Line P1P2, or "on" if this point is lying on line P1P2, or "right" if this is on the right side of Line P1P2.
Sample Input
141 1 1 2 3 3 2 1 1 3
Sample output
OnLeftOnRight
Idea: Cross Product
<span style="font-size:18px;">#include <cstdio>#include <iostream>using namespace std;#define N 1005double a[N],b[N];int main(){int t;double x1,x2,y1,y2,sum;scanf("%d",&t);while(t--){int n,i;scanf("%d",&n);for(i = 1; i <= n; i++)scanf("%lf%lf",&a[i],&b[i]);int p1,p2; scanf("%d%d",&p1,&p2); x1 = a[p2] - a[p1]; y1 = b[p2] - b[p1];for(i=1; i<=n; i++){ x2 = a[i] - a[p1]; y2 = b[i] - b[p1]; sum = x1*y2 - x2*y1;if(sum < 0)printf("Right\n");else if(sum > 0)printf("Left\n");else printf("On\n"); }}return 0;}</span>
