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题目
{% fold 点击显/隐题目 %}
Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick on top of them. Stan has noticed that the last thrown stick is always on top but he wants to know all the sticks that are on top. Stan sticks are very, very thin such that their thickness can be neglected.
Input consists of a number of cases. The data for each case start with 1 <= n <= 100000, the number of sticks for this case. The following n lines contain four numbers each, these numbers are the planar coordinates of the endpoints of one stick. The sticks are listed in the order in which Stan has thrown them. You may assume that there are no more than 1000 top sticks. The input is ended by the case with n=0. This case should not be processed.
For each input case, print one line of output listing the top sticks in the format given in the sample. The top sticks should be listed in order in which they were thrown.
The picture to the right below illustrates the first case from input.
5 1 1 4 2 2 3 3 1 1 -2.0 8 4 1 4 8 2 3 3 6 -2.0 3 0 0 1 1 1 0 2 1 2 0 3 1 0
Top sticks: 2, 4, 5. Top sticks: 1, 2, 3.
题解
按顺序扔木棍,找到所有上面没有压东西的木棍
基本思路就是判断线段相交
不过看数据量O(n2)会爆炸
然而搜了一下大家全都这样写的……
交了一次 TLE
对了一下发现对for的循环顺序还有要求……
不懂出题人的高级操作
代码
{% fold 点击显/隐代码 %}```cpp Pick-up sticks 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;
typedef complex Vector;
typedef Vector Point;
struct Segment {
Point a, b;
Segment() {}
Segment(Point _a, Point _b) : a(_a), b(_b) {}
Vector toVector() { return b - a; }
};
inline Point read_Point() {
double x, y;
scanf("%lf%lf", &x, &y);
return Point(x, y);
}
inline int sgn(const double &x) {
if (fabs(x) < eps)
return 0;
return x > 0 ? 1 : -1;
}
inline bool operator==(const Vector &vec1, const Vector &vec2) {
return sgn(vec1.real() - vec2.real()) == 0 &&
sgn(vec1.imag() - vec2.imag()) == 0;
}
inline bool operator!=(const Vector &vec1, const Vector &vec2) {
return !(vec1 == vec2);
}
inline int Cross(const Vector &a, const Vector &b) {
return a.real() * b.imag() - a.imag() * b.real();
}
inline int Dot(const Vector &a, const Vector &b) {
return a.real() * b.real() + a.imag() * b.imag();
}
inline double Distance(const Point &a, const Point &b) { return abs(a - b); }
inline int Point_Segment(const Vector &p, const Segment &L) {
return sgn(Cross(L.b - L.a, p - L.a));
}
inline int Segment_Segment(const Segment L1, const Segment L2) {
double a = L1.b.real() - L1.a.real();
double b = L2.b.real() - L2.a.real();
double c = L1.b.imag() - L1.a.imag();
double d = L2.b.imag() - L2.a.imag();
double f = a * d - b * c;
// 平行或重叠
if (sgn(f) == 0)
return -1;
double g = L2.b.real() - L1.a.real();
double h = L2.b.imag() - L1.a.imag();
double t = (d * g - b * h) / f;
double s = (-c * g + a * h) / f;
// 在延长线上
if (t < 0 || t > 1 || s < 0 || s > 1)
return 0;
// 线段相交
return 1;
}
const int maxn = 100005;
bool flag[maxn];
Segment Segments[maxn];
int main() {
int n;
while (scanf("%d", &n), n) {
for (int i = 0; i < n; ++i) {
Point p1 = read_Point();
Point p2 = read_Point();
Segments[i] = Segment(p1, p2);
}
memset(flag, false, sizeof(bool) * (n + 5));
for (int i = 0; i < n; ++i) {
for (int j = i + 1; j < n; ++j) {
if (Segment_Segment(Segments[i], Segments[j]) > 0) {
flag[i] = true;
break;
}
}
}
bool first = true;
printf("Top sticks:");
for (int i = 0; i < n; ++i) {
if (!flag[i]) {
if (first) {
first = false;
printf(" ");
} else {
printf(", ");
}
printf("%d", i + 1);
}
}
printf(".\n");
}
return 0;
}
{% endfold %}