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={(oldsymbol{x_1},y_1),(oldsymbol{x_2},y_2),...,(oldsymbol{x_N},y_N)}), with their inputs (oldsymbol{x}in mathbb{R}^d), and outputs (yin {−1,1}). We will try to find a function (f(oldsymbol{x})=sign(sum_{i=1}^d w_icdot x_i+b)=sign(oldsymbol{w^T} cdot oldsymbol{x}+b)) so that (f(oldsymbol{x_i})=y_i,i=1,2,...,N).
(oldsymbol{w}, oldsymbol{x}) mentioned above are all d-dimensional vectors, i.e. (oldsymbol{w}=(w_1,w_2,...,w_d), oldsymbol{x}=(x_1,x_2,...,x_d)). To simplify the question, let (w_0=b), (x_0=1), then (f(oldsymbol{x})=sign(sum_{i = 0}^d w_icdot x_i)=sign(oldsymbol{w^T}cdot oldsymbol{x})). Therefore, finding a satisfying function (f(oldsymbol{x})) is equivalent to finding a proper (oldsymbol{w}).
To solve the problem, we have a algorithm, PLA(Popcorn Label Algorithm).
Accoding to PLA, we will randomly generate (oldsymbol{w}).
If (f(oldsymbol{x})=sign(oldsymbol{w^T}cdot oldsymbol{x})) fails to give any element ((oldsymbol{x_i},y_i)in D) the right classification, i.e. (f(oldsymbol{x_i}) eq 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 wBut 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)= egin{cases} -1& x < 0 \ 0& x = 0 \ 1& x > 0 end{cases} ]
Input
The first line contains an integer (T(1le Tle 1000)), the number of test cases.
Each test case begins with a line containing a single integer (n(1le nle 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 (−10^5le x_{i,1},x_{i,2}le 10^5, y_iin {−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
3
2
1 1 1
2 0 -1
4
0 0 1
2 0 -1
1 1 1
1 -1 -1
6
0 0 1
2 0 -1
1 1 1
1 -1 -1
1 0 1
0 1 -1
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!
");
return;
}
}
for(int i = 0; i < c2; ++i) {
if(isPointInPolygon(s2[i], s1)) {
printf("Infinite loop!
");
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!
");
return;
}
}
}
printf("Successful!
");
}
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!
");
continue;
}
if(s1.size()) s1 = Andrew(s1);
if(s2.size()) s2 = Andrew(s2);
solve(s1, s2);
}
return 0;
}