Hihocoder 1275 扫地机器人 计算几何
题意:
有一个房间的形状是多边形,而且每条边都平行于坐标轴,按顺时针给出多边形的顶点坐标
还有一个正方形的扫地机器人,机器人只可以上下左右移动,不可以旋转
问机器人移动的区域能不能覆盖整个房间
分析:
#include <cstdio>
#include <cmath>
#include <algorithm>
using std::abs;
using std::swap;
const int maxn = 1010;
struct Point
{
int x, y;
Point(int x = 0, int y = 0): x(x), y(y) {}
void read() { scanf("%d%d", &x, &y); }
Point operator - (const Point& t) const {
return Point(x - t.x, y - t.y);
}
};
int n, m;
Point p[maxn], dir[maxn]; //dir是边的单位方向向量
int angle[maxn]; //用叉积判断内外角,angle为正是外角,为负是内角
int sign(int x) { if(!x) return 0; return x > 0 ? 1 : -1; }
Point Normalize(Point A) {
return Point(sign(A.x), sign(A.y));
}
int Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
int Length(Point A) { if(A.x) return abs(A.x); return abs(A.y); }
int left, right, top, bottom; //机器人沿某条边扫过的矩形
//判断扫描区域是否与房间的某条边相交
bool intersect() {
for(int i = 1; i <= n; i++) {
if(p[i].x <= left && p[i+1].x <= left) continue;
if(p[i].x >= right && p[i+1].x >= right) continue;
if(p[i].y <= bottom && p[i+1].y <= bottom) continue;
if(p[i].y >= top && p[i+1].y >= top) continue;
return true;
}
return false;
}
int main()
{
int q; scanf("%d", &q);
while(q--) {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) p[i].read();
p[n + 1] = p[1];
for(int i = 1; i <= n; i++) dir[i] = Normalize(p[i+1]-p[i]);
dir[0] = dir[n];
for(int i = 1; i <= n; i++) angle[i] = Cross(dir[i - 1], dir[i]);
angle[n + 1] = angle[1];
bool ok = true;
for(int i = 1; i <= n; i++) {
if(angle[i] < 0 && angle[i+1] < 0 && Length(p[i+1]-p[i]) < m) {
ok = false; break;
}
if(dir[i].x) {
int delta = m * dir[i].x;
left = p[i].x; right = p[i+1].x;
if(angle[i] > 0) left -= delta;
if(angle[i+1] > 0) right += delta;
top = p[i].y; bottom = top - delta;
} else {
int delta = m * dir[i].y;
top = p[i].y; bottom = p[i+1].y;
if(angle[i] > 0) top -= delta;
if(angle[i+1] > 0) bottom += delta;
left = p[i].x; right = left + delta;
}
if(left > right) swap(left, right);
if(bottom > top) swap(bottom, top);
if(intersect()) { ok = false; break; }
}
printf("%s\n", ok ? "Yes" : "No");
}
return 0;
}