0%

UVA 10256 The Great Divide (判断凸包相交)

题目链接:UVA 10256

题意

有n个红点和m个蓝点,问是否存在一条直线能够分开红点和蓝点。

题解

分别求出红点和蓝点的凸包,判断两个凸包是否相交。

凸包不相交的条件:

  • 凸包上的任意点都在另一个凸包的外面
  • 凸包的任意线段都与另一个凸包不相交

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
#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("No\n");
return;
}
}
for(int i = 0; i < c2; ++i) {
if(isPointInPolygon(s2[i], s1)) {
printf("No\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("No\n");
return;
}
}
}
printf("Yes\n");
}

int main() {
int n, m;
while (cin >> n >> m) {
if(n == 0 && m == 0) break;
Polygon s1, s2;
for (int i = 0; i < n; ++i) {
double x1, x2;
scanf("%lf%lf", &x1, &x2);
s1.push_back(Point(x1, x2));
}
for (int i = 0; i < m; ++i) {
double x1, x2;
scanf("%lf%lf", &x1, &x2);
s2.push_back(Point(x1, x2));
}
if(s1.size()) s1 = Andrew(s1);
if(s2.size()) s2 = Andrew(s2);
solve(s1, s2);
}
return 0;
}

参考

《算法竞赛入门经典》 刘汝佳 / 陈锋

欢迎关注我的其它发布渠道