POJ 2318 叉积 + 枚举

code:

#include 
#include 
using namespace std;

const int N = 5000 + 10;

int Ui[N], Li[N], ans[N];
int n, m, x1, y1, x2, y2, up, down, x, y;

bool check(int x, int y, int ux, int uy, int dx, int dy){
	return (ux - x) * (dy - y) - (dx - x) * (uy - y) > 0;
}

int main(){

	bool f = 0;
	while(cin >> n, n){
		memset(ans, 0, sizeof ans);
		cin >> m >> x1 >> y1 >> x2 >> y2;
		if(f) cout << endl;
		else f = 1;
		for(int i = 1; i <= n; ++i) cin >> Ui[i] >> Li[i];
		Ui[0] = Li[0] = x1;
		Ui[n + 1] = Li[n + 1] = x2;
		for(int i = 1; i <= m; ++i){
			cin >> x >> y;
			for(int j = 0; j <= n; ++j){
				if(check(x, y, Ui[j], y1, Li[j], y2) && !check(x, y, Ui[j + 1], y1, Li[j + 1], y2)){
					++ans[j];
					break;
				}
			}
		}
		for(int i = 0; i <= n; ++i) cout << i << ": " << ans[i] << endl;
	}

	return 0;
}