和POJ2318一样的方法,都是利用叉积判断+二分,不过这题要先排序,还有输出的是,每个数量的格子数
代码:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1005;
int n, m, x1, y1, x2, y2;
struct Point {
int x, y;
Point() {}
Point(int x, int y) {
this->x = x;
this->y = y;
}
};
typedef Point Vector;
Vector operator - (Vector A, Vector B) {
return Vector(A.x - B.x, A.y - B.y);
}
struct Seg......
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