题目

{% fold 点击显/隐题目 %}

题解

对一个只能向左转的蚂蚁,求它尽可能经过最多点的路径

容易画出来蚂蚁必然能通过所有点

本质上就是将整个散点图分成一层层凸包

需要特别注意除了最外层凸包,里面的凸包起点要由上一层凸包决定

代码

{% fold 点击显/隐代码 %}```cpp Space Ant https://github.com/OhYee/sourcecode/tree/master/ACM 代码备份
#include
#include
#include
#include
#include

using namespace std;
#define Log(format, ...) // printf(format, ##VA_ARGS)

/* 计算几何模板 */

const double eps = 1e-8;
const double INF = 9e50;
inline int sgn(const double &x) {
if (fabs(x) < eps)
return 0;
return x > 0 ? 1 : -1;
}

struct Vector {
double x, y;
int n;
Vector(double _x = 0, double _y = 0, int _n = 0) : x(_x), y(_y), n(_n) {}
bool operator==(const Vector &rhs) const {
return sgn(x - rhs.x) == 0 && sgn(y - rhs.y) == 0;
}
bool operator!=(const Vector &rhs) const { return !(*this == rhs); }
bool operator<(const Vector &rhs) const {
if (sgn(y - rhs.y) == 0)
return sgn(x - rhs.x) < 0;
return sgn(y - rhs.y) < 0;
}
Vector operator+(const Vector &rhs) const {
return Vector(x + rhs.x, y + rhs.y);
}
Vector operator-(const Vector &rhs) const {
return Vector(x - rhs.x, y - rhs.y);
}
double squre() const { return x * x + y * y; }
double distance() const { return sqrt(squre()); }
void print(bool flag = 0) const {
Log("(%.f %.f)", x, y);
if (flag)
Log("\n");
}
};
typedef Vector Point;
struct Segment {
Point a, b;
Segment() {}
Segment(Point _a, Point _b) : a(_a), b(_b) {}
Vector toVector() const { return b - a; }
double distance() const { return (b - a).distance(); }
void print(bool flag = 0) const {
a.print();
Log(" -> ");
b.print();
if (flag)
Log("\n");
}
};

/**
* 读入一个点的坐标
* @return 读入的点
*/
inline Point read_Point() {
double x, y;
scanf("%lf%lf", &x, &y);
return Point(x, y);
}

/**
* 计算两个向量的叉积
* @param a 向量1
* @param b 向量2
* @return 叉积
*/
inline double Cross(const Vector &a, const Vector &b) {
return a.x * b.y - a.y * b.x;
}

/**
* 计算两个向量的点积
* @param a 向量1
* @param b 向量2
* @return 点积
*/
inline double Dot(const Vector &a, const Vector &b) {
return a.x * b.x + a.y * b.y;
}

/**
* 计算两点之间的距离
* @param a 线段L1
* @param b 线段L2
* @return 两点间的距离
*/
inline double Distance(const Point &a, const Point &b) {
return (a - b).distance();
}

/**
* 点和直线的关系
* @param p 目标点
* @param L 目标直线
* @return 1 在左侧,0 在直线上,-1在右侧
*/
inline int Point_Segment(const Vector &p, const Segment &L) {
return sgn(Cross(L.b - L.a, p - L.a));
}

/**
* 计算两个线段的位置关系
* @param L1 线段L1
* @param L2 线段L2
* @param p 返回交点坐标
* @return 2 重叠
1 相交
0 延长线相交
-1 平行
-2 共线不交
*/
inline int Segment_Segment(const Segment L1, const Segment L2,
Point *p = NULL) {
double a = L1.b.x - L1.a.x;
double b = L2.b.x - L2.a.x;
double c = L1.b.y - L1.a.y;
double d = L2.b.y - L2.a.y;
double f = a * d - b * c;
// 平行或重叠
if (sgn(f) == 0) {
if (Point_Segment(L1.a, L2)) {
// 平行
return -1;
} else {
// 共线
int len = max(max(Distance(L1.a, L2.a), Distance(L1.a, L2.b)),
max(Distance(L1.b, L2.a), Distance(L1.b, L2.b)));
if (sgn(len - L1.distance() - L2.distance()) > 0) {
// 共线不交
return -2;
} else {
// 重叠
return 2;
}
}
}
double g = L2.b.x - L1.a.x;
double h = L2.b.y - L1.a.y;
double t = (d * g - b * h) / f;
double s = (-c * g + a * h) / f;
if (p != NULL)
*p = Point(L1.a.x + t * a, L1.a.y + t * c);
// 在延长线上
if (t < 0 || t > 1 || s < 0 || s > 1)
return 0;
// 线段相交
return 1;
}

/**
* 判断点是否在多边形内部
* @param p 需要判断的点
* @param polygon 多边形点集,需要保证有序
* @param numberOfSide 多边形边数
* @return true 点在多边形内,false 点不在多边形内
*/
bool Point_Polygon(Point p, Point polygon[], int numberOfSide) {
bool ok =
Point_Segment(p, Segment(polygon[numberOfSide - 1], polygon[0])) >= 0;
for (int i = 1; i < numberOfSide && ok; ++i) {
if (!(Point_Segment(p, Segment(polygon[i - 1], polygon[i])) >= 0))
ok = false;
}
return ok;
}

/**
* 求点集的凸包
* @param p 点集,需要保证已经排序
* @param numOfPoint 点集内的点的个数
* @param vis 记录点对应的编号是否可以选择
* @param ans 返回的凸包
* @param begin 起始位置
* @return 返回凸包上点的个数
*/
int Convex_Hull(Point p[], int numOfPoint, bool vis[], Point ans[],
int begin = 0) {
int pos = begin - 1;
// 右链
for (int i = 0; i < numOfPoint; ++i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
// 左链
for (int i = numOfPoint - 2; i > 0; --i) {
if (!vis[p[i].n]) {
while (pos > 0 &&
Point_Segment(p[i], Segment(ans[pos], ans[pos - 1])) > 0) {
vis[ans[pos--].n] = false;
}
ans[++pos] = p[i];
vis[ans[pos].n] = true;
}
}
return pos + 1;
}

const int maxn = 105;

int n;
Point points[maxn];
Point ans[maxn];
bool vis[maxn];

void solve() {
int len = 0;
sort(points, points + n);
memset(vis, false, (n + 5) * sizeof(bool));

while (len != n) {
    len = Convex_Hull(points, n, vis, ans, len);
    Log("%d\n", len);
}

printf("%d", len);
for (int i = 0; i < len; ++i)
    printf(" %d", ans[i].n);
printf("\n");

}

int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
int t;
scanf("%d", &t);
points[i] = read_Point();
points[i].n = t;
}
solve();
}
return 0;
}

{% endfold %}