HDU 6590 Code (判断凸包相交)

2019 杭电多校 1 1013

题目链接：HDU 6590

比赛链接：2019 Multi-University Training Contest 1

Problem Description

After returning with honour from ICPC(International Cat Programming Contest) World Finals, Tom decides to say goodbye to ICPC and start a new period of life. He quickly gets interested in AI.

In the subject of Machine Learning, there is a classical classification model called perceptron, defined as follows:

Assuming we get a set of training samples: \(D={(\boldsymbol{x_1},y_1),(\boldsymbol{x_2},y_2),...,(\boldsymbol{x_N},y_N)}\), with their inputs \(\boldsymbol{x}\in \mathbb{R}^d\), and outputs \(y\in \{−1,1\}\). We will try to find a function \(f(\boldsymbol{x})=sign(\sum_{i=1}^d w_i\cdot x_i+b)=sign(\boldsymbol{w^T} \cdot \boldsymbol{x}+b)\) so that \(f(\boldsymbol{x_i})=y_i,i=1,2,...,N\).

\(\boldsymbol{w}, \boldsymbol{x}\) mentioned above are all d-dimensional vectors, i.e. \(\boldsymbol{w}=(w_1,w_2,...,w_d), \boldsymbol{x}=(x_1,x_2,...,x_d)\). To simplify the question, let \(w_0=b\), \(x_0=1\), then \(f(\boldsymbol{x})=sign(\sum_{i = 0}^d w_i\cdot x_i)=sign(\boldsymbol{w^T}\cdot \boldsymbol{x})\). Therefore, finding a satisfying function \(f(\boldsymbol{x})\) is equivalent to finding a proper \(\boldsymbol{w}\).

To solve the problem, we have a algorithm, PLA(Popcorn Label Algorithm).

Accoding to PLA, we will randomly generate \(\boldsymbol{w}\).

If \(f(\boldsymbol{x})=sign(\boldsymbol{w^T}\cdot \boldsymbol{x})\) fails to give any element \((\boldsymbol{x_i},y_i)\in D\) the right classification, i.e. \(f(\boldsymbol{x_i})\neq y_i\), then we will replace \(w\) with another random vector. We will do this repeatedly until all the samples \(\in D\) are correctly classified.

As a former-JBer, Tom excels in programming and quickly wrote the pseudocode of PLA.
  w := a random vector

  while true do

      flag:=true

      for i:=1 to N do

          if f(x[ i ]) != y[ i ] then

              flag:=false

              break

      if flag then

          break

      else

          w := a random vector

  return w
But Tom found that, in some occasions, PLA will end up into an infinite loop, which confuses him a lot. You are required to help Tom determine, when performed on a given sample set \(D\), if PLA will end up into an infinite loop. Print Infinite loop! if so, or Successful! otherwise.

We only consider cases when \(d=2\) for simplification.

Note:

\[sign(x)= \begin{cases} -1& x < 0 \\ 0& x = 0 \\ 1& x > 0 \end{cases}
\]

Input

The first line contains an integer \(T(1\le T\le 1000)\), the number of test cases.

Each test case begins with a line containing a single integer \(n(1\le n\le 100)\), size of the set of training samples \(D\).

Then \(n\) lines follow, the ith of which contains three integers \(x_{i,1},x_{i,2},y_i (−10^5\le x_{i,1},x_{i,2}\le 10^5, y_i\in {−1,1})\), indicating the ith sample \((x_i,y_i)\) in \(D\), where \(x_i=(x_{i,1},x_{i,2})\).

Output

For each test case, output a single line containing the answer: “Infinite loop!” or “Successful!”.

Sample Input

Sample Output

Successful!

Successful!

Infinite loop!

Solution

题意

给出两类点的坐标，问能否用一条直线将两类点分开。

题解

题目看懂了就很好做了。

就是分别对两类点求凸包，然后判断两个凸包是否相交。若不相交，则能够用一条直线分开两类点，否则不能。

其实就是判断凸包是否相交的模板题。

类似的题目有：

Code

#include <bits/stdc++.h>

using namespace std;

const double eps = 1e-8;

const double pi = acos(-1.0);

class Point {

public:

    double x, y;

    Point(double x = 0, double y = 0) : x(x), y(y) {}

    Point operator+(Point a) {

        return Point(a.x + x, a.y + y);

    }

    Point operator-(Point a) {

        return Point(x - a.x, y - a.y);

    }

    bool operator<(const Point &a) const {

        if (x == a.x)

            return y < a.y;

        return x < a.x;

    }

    bool operator==(const Point &a) const {

        if (fabs(x - a.x) < eps && fabs(y - a.y) < eps)

            return 1;

        return 0;

    }

    double length() {

        return sqrt(x * x + y * y);

    }

};

typedef Point Vector;

double cross(Vector a, Vector b) {

    return a.x * b.y - a.y * b.x;

}

double dot(Vector a, Vector b) {

    return a.x * b.x + a.y * b.y;

}

bool isclock(Point p0, Point p1, Point p2) {

    Vector a = p1 - p0;

    Vector b = p2 - p0;

    if (cross(a, b) < -eps)

        return true;

    return false;

}

double getDistance(Point a, Point b) {

    return sqrt(pow(a.x - b.x, 2) + pow(a.y - b.y, 2));

}

typedef vector<Point> Polygon;

Polygon Andrew(Polygon s) {

    Polygon u, l;

    if(s.size() < 3) return s;

    sort(s.begin(), s.end());

    u.push_back(s[0]);

    u.push_back(s[1]);

    l.push_back(s[s.size() - 1]);

    l.push_back(s[s.size() - 2]);

    for(int i = 2 ; i < s.size() ; ++i) {

        for(int n = u.size() ; n >= 2 && !isclock(u[n - 2], u[n - 1], s[i]); --n) {

            u.pop_back();

        }

        u.push_back(s[i]);

    }

    for(int i = s.size() - 3 ; i >= 0 ; --i) {

        for(int n = l.size() ; n >=2 && !isclock(l[n-2],l[n-1],s[i]); --n) {

            l.pop_back();

        }

        l.push_back(s[i]);

    }

    for(int i = 1 ; i < u.size() - 1 ; i++) l.push_back(u[i]);

    return l;

}

int dcmp(double x)  {

    if (fabs(x) <= eps)

        return 0;

    return x > 0 ? 1 : -1;

}

// 判断点在线段上

bool OnSegment(Point p, Point a1, Point a2) {

    return dcmp(cross(a1 - p, a2 - p)) == 0 && dcmp(dot(a1 - p, a2 - p)) < 0;

}

// 判断线段相交

bool Intersection(Point a1, Point a2, Point b1, Point b2) {

    double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1),

            c3 = cross(b2 - b1, a1 - b1), c4 = cross(b2 - b1, a2 - b1);

    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;

}

// 判断点在凸包内

int isPointInPolygon(Point p, vector<Point> s) {

    int wn = 0, cc = s.size();

    for (int i = 0; i < cc; i++) {

        Point p1 = s[i];

        Point p2 = s[(i + 1) % cc];

        if (p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1;

        int k = dcmp(cross(p2 - p1, p - p1));

        int d1 = dcmp(p1.y - p.y);

        int d2 = dcmp(p2.y - p.y);

        if (k > 0 && d1 <= 0 && d2 > 0) wn++;

        if (k < 0 && d2 <= 0 && d1 > 0) wn--;

    }

    if (wn != 0) return 1;

    return 0;

}

void solve(Polygon s1, Polygon s2) {

    int c1 = s1.size(), c2 = s2.size();

    for(int i = 0; i < c1; ++i) {

        if(isPointInPolygon(s1[i], s2)) {

            printf("Infinite loop!\n");

            return;

        }

    }

    for(int i = 0; i < c2; ++i) {

        if(isPointInPolygon(s2[i], s1)) {

            printf("Infinite loop!\n");

            return;

        }

    }

    for (int i = 0; i < c1; i++) {

        for (int j = 0; j < c2; j++) {

            if (Intersection(s1[i], s1[(i + 1) % c1], s2[j], s2[(j + 1) % c2])) {

                printf("Infinite loop!\n");

                return;

            }

        }

    }

    printf("Successful!\n");

}

int main() {

    int T;

    cin >> T;

    while (T--) {

        int n;

        scanf("%d", &n);

        Polygon s1, s2;

        for (int i = 0; i < n; ++i) {

            double x1, x2, y;

            scanf("%lf%lf%lf", &x1, &x2, &y);

            if(y == 1) {

                s1.push_back(Point(x1, x2));

            } else {

                s2.push_back(Point(x1, x2));

            }

        }

        if(n == 1) {

            printf("Successful!\n");

            continue;

        }

        if(s1.size()) s1 = Andrew(s1);

        if(s2.size()) s2 = Andrew(s2);

        solve(s1, s2);

    }

    return 0;

}

巴特西

HDU 6590 Code (判断凸包相交)

Problem Description

Input

Output

Sample Input

Sample Output

Solution

题意

题解

Code

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